UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-28T20:58:27Z

SabrinaPannu:

==Logarithmic Scale==
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| LOGARITHMIC SCALE
|-
|style="background: #FBBBB9; padding:12px;"|
A '''Logarithmic Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithm scale.
|}
==Richter Magnitude Scale==

[[File:Rc.png]]

Invented by Charles Richter in 1935, the Richter Scale or the
'''Local Magnitude Scale (ML)''' measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake.

The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is '''tenfold.''' This means that each number increase on the Richter scale indicates an intensity ten times stronger. The range of the Richter scale is between 0 and 10. An earthquake can measure above 10.0, which is then called an epic earthquake.

An earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is '''10 x 10 = 100''' times stronger than an earthquake of magnitude 5. An earthquake of magnitude 8 is ''' 10 x 10 x 10 = 1000''' times stronger than an earthquake of magnitude 5.

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Comparing Earthquakes'''
|-
|style="background: #ADDFFF; padding:12px;"|

How does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
*The increase in magnitude is 5-2 or 3.

*<math>10^3 = 1000</math>; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

How does a magnitude '''7''' quake compare to a magnitude '''1''' earthquake?
*The increase in magnitude is 7-1 or 6.
*<math>10^6=1 000 000</math> thus the ground movement of a magnitude 7 quake is '''one million times greater''' than a magnitude 1 quake.

'''
As illustrated in the examples above, there is are large differences between the impact of earthquakes because of the Richter scale being a logarithmic scale.'''

|}
[[File:Richterscale.png]]
----

The Richter magnitude of an earthquake is determined from the
logarithm of the amplitude of waves recorded by seismographs. </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref> The original formula is: [[File:Log.png]]

Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by '''"R= log[ I ÷ I0]"'''
*I0 is the "threshold quake", or movement that can barely be detected and I, the intensity, is given in terms of multiples of that threshold intensity.
*Suppose a seismograph measures that the I= 989I0.
**To determine the magnitude of the earthquake, the intensity to a Richter rating by evaluating the Richter function at '''I= 98910:R = log[I ÷ I0 ]=log[ 98910 ÷ I0] =log[989]=2.99519629'''... or about 3 on the Richter Scale, which turns out to be a minor earthquake.
----

{{#ev:youtube | tcmrlR2XMNM| 400}}

----

''' References:''' <references/>

Course:MATH110/Archive/2010-2011/003/Teams/Valais

2011-01-28T20:53:57Z

SabrinaPannu: /* Homework */

{{
Infobox MATH110 Teams
| team name = Valais
| member 1 = Albert König
| member 2 = Charlene Ng
| member 3 = Jonathan Rothwell
| member 4 = [[User:SabrinaPannu|Sabrina Pannu]]
}}
In workshop G.
==Contact Information==
* Sabrina Pannu : # 604-786-1757 ''email:'' pannu7@msn.com
'''Everyone please add your contact information!''' [[User:SabrinaPannu|SabrinaPannu]]

==Homework==
[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_11 Homework 11]

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_12 Homework 12]

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_13 Homework 13]

==Key Words==
* Sabrina - sine

* Jonathan - tangent

* Albert - cotangent

* Charlene - cosine

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-28T20:53:11Z

SabrinaPannu:

==Logarithmic Scale==
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| LOGARITHMIC SCALE
|-
|style="background: #FBBBB9; padding:12px;"|
A '''Logarithmic Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithm scale.
|}
==Richter Magnitude Scale==

[[File:Rc.png]]

Invented by Charles Richter in 1935, the Richter Scale or the
'''Local Magnitude Scale (ML)''' measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake.

The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is '''tenfold.''' This means that each number increase on the Richter scale indicates an intensity ten times stronger. The range of the Richter scale is between 0 and 10. An earthquake can measure above 10.0, which is then called an epic earthquake.

An earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is '''10 x 10 = 100''' times stronger than an earthquake of magnitude 5. An earthquake of magnitude 8 is ''' 10 x 10 x 10 = 1000''' times stronger than an earthquake of magnitude 5.

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Comparing Earthquakes'''
|-
|style="background: #ADDFFF; padding:12px;"|

How does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
*The increase in magnitude is 5-2 or 3.

*<math>10^3 = 1000</math>; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

How does a magnitude '''7''' quake compare to a magnitude '''1''' earthquake?
*The increase in magnitude is 7-1 or 6.
*<math>10^6=1 000 000</math> thus the ground movement of a magnitude 7 quake is '''one million times greater''' than a magnitude 1 quake.

'''
As illustrated in the examples above, there is are large differences between the impact of earthquakes because of the Richter scale being a logarithmic scale.'''

|}
[[File:Richterscale.png]]

The Richter magnitude of an earthquake is determined from the
logarithm of the amplitude of waves recorded by seismographs. </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref> The original formula is: [[File:Log.png]]

Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by '''"R= log[ I ÷ I0]"'''
*I0 is the "threshold quake", or movement that can barely be detected and I, the intensity, is given in terms of multiples of that threshold intensity.
*Suppose a seismograph measures that the I= 989I0.
**To determine the magnitude of the earthquake, the intensity to a Richter rating by evaluating the Richter function at '''I= 98910:R = log[I ÷ I0 ]=log[ 98910 ÷ I0] =log[989]=2.99519629'''... or about 3 on the Richter Scale, which turns out to be a minor earthquake.

{{#ev:youtube | tcmrlR2XMNM| 400}}

''' References:''' <references/>

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-28T20:16:41Z

SabrinaPannu:

==Logarithmic Scale==

A '''Logarithmic Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithm scale.

==Richter Magnitude Scale==
Invented by Charles Richter in 1935, the Richter Scale or the
'''Local Magnitude Scale (ML)''' measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake.

[[File:Rc.png]]

The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is '''tenfold.''' This means that each number increase on the Richter scale indicates an intensity ten times stronger. The range of the Richter scale is between 0 and 10. An earthquake can measure above 10.0, which is then called an epic earthquake.

An earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is '''10 x 10 = 100''' times stronger than an earthquake of magnitude 5. An earthquake of magnitude 8 is ''' 10 x 10 x 10 = 1000''' times stronger than an earthquake of magnitude 5.

Example"
A magnitude 2 quake causes 10 times more ground movement than a magnitude 1 quake.

How does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
The increase in magnitude is 5-2 or 3.

10^3 = 1000; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

As illustrated in the examples above, there is are large differences between the magnitudes of earthquakes because of the Richter scale being a logarithmic scale.
The Richter magnitude of an earthquake is determined from the logarithm of the amplitude of waves recorded by seismographs.

[[File:Richterscale.png]]

Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by '''"R= log[ I ÷ I0]"''' {I0 is the "threshold quake", or movement that can barely be detected and I, the intensity, is given in terms of multiples of that threshold intensity. Suppose a seismograph measures that the I= 989I0. To determine the magnitude of the earthquake, the intensity to a Richter rating by evaluating the Richter function at '''I= 98910:R = log[I ÷ I0 ]=log[ 98910 ÷ I0] =log[989]=2.99519629'''... or about 3 on the Richter Scale, which turns out to be a minor earthquake.

The Richter magnitude of an earthquake is determined from the
logarithm of the amplitude of waves recorded by seismographs. </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref> The original formula is: [[File:Log.png]]

''' References:''' <references/>

File:Log.png

2011-01-28T20:11:53Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-28T20:08:51Z

SabrinaPannu:

File:Rc.png

2011-01-28T19:44:29Z

SabrinaPannu:

File:Richterscale.png

2011-01-28T19:40:02Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 13

2011-01-28T19:27:14Z

SabrinaPannu: Created page with "==Logarithm Scale== A '''Logarithm Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale n..."

==Logarithm Scale==

A '''Logarithm Scale''' can be defined as the scale on which actual distances from the origin are proportional to the logarithms of the corresponding scale numbers <ref> Princeton, 2010 http://wordnetweb.princeton.edu/perl/webwn?s=logarithmic%20scale </ref> or scale of measurement that uses the logarithm of a physical quantity instead of the quantity itself <ref> Wikipedia, 2011 http://en.wikipedia.org/wiki/Logarithmic_scale </ref>
A log scale makes it more helpful and manageable to compare a large range of values.
The logarithmic scale can be helpful when the data covers a large range of values – the logarithm reduces this to a more manageable range. For example, a graph's axis is labelled 10, 100, 1000, and 100000 instead of 1, 2, 3 and 4. This is a common and useful measurement scale that is used for measuring entropy, pH, acoustic power, entropy and sesimic energy among many other things. We will investigate how the Richter Magnitude Scale works on a logarithim scale.

==Richter Magnitude Scale==
Invented by Charles Richter in 1934, the Richter Scale measures the magnitude of earthquakes. More precisely, the Richter scale is a measure of ground movement at the epicenter of an earthquake which can be in turn be used to determine the sesimic energy released by an earthquake. The Richter Scale is a logarithmic scale; thus, each order of magnitude increase is tenfold.

Although it may seem that there is little difference between a magnitude 1 quake and magnitude 2 quake, there is actually an immense difference in the power of these two earthquakes.
A magnitude 2 quake causes 10 times more ground movement than a magnitude 1 quake
So how does a magnitude '''5''' quake compare to a magnitude '''2''' earthquake?
The increase in magnitude is 5-2 or 3.

10^3 = 1000; thus the ground movement of a magnitude 5 quake is '''1000 times greater''' than a magnitude 2 quake.

As illustrated in the examples above, there is are large differences between the magnitudes of earthquakes because of the Richter scale being a logarithmic scale.

Charles Richter defined the magnitude of the earthquake as the following:

M=log I/S

<ref>SOS Math http://www.sosmath.com/algebra/logs/log5/log56/log56.html</ref>

''' References:''' <references/>

User:SabrinaPannu

2011-01-28T08:25:16Z

SabrinaPannu:

Hi, my name is Sabrina and I'm in the Faculty of Arts :)

 The Pythagorean Theorem 

The Pythagorean Theorem was discovered by the famous Greek mathematician and philosopher Pythagoras. However, there is some controversy in the mathematical community as some suggest Pythagoras may have stolen his famous formula from the Babylonians. The theorem is commonly written as an equation, known as the Pythagorean equation: a2 + b2 = c2. In this simplistic formula, the letters refer to the three sides of a right-angled triangle with a and b signifying the sides of the triangle and c representing the hypotenuse. Essentially, the Pythagorean Theorem allows us to find the missing length of the side of a right triangle and is taught in geometry classes globally as it has extraordinary relevance to our daily lives. It is widely used in the field of engineering. The Pythagorean Theorem also has hundreds of proofs giving it immense credibility; in fact, it has more proofs than any other mathematical theorem in the world.

Sources: Wikipedia-Pythagorean Thereom[http://en.wikipedia.org/wiki/Pythagorean_theorem ]

==Calculus in Economics==

[[File:Economics2.jpg]]

Calculus is the branch of mathematics concerning the identification and properties of derivatives and integrals of functions and has many practical applications in many fields. One of these fields is economics. Economics and mathematics are closely related as calculus is essentially the language of economics. Calculus is applied in even the basics of microeconomics when firms are trying to maximize their profit, or when individuals are looking to increase utility. A firm produces outputs through the basic principle that marginal cost = marginal revenue. The term “marginal”in economics can be thought of as synonymous with the “derivative of ” as in calculus. The profit is revenue minus cost and in order to maximize it, the equation must be set to zero. Therefore profit is maximized when both marginal cost and marginal revenue are equal to each other. Calculus aids business owners to figure out marginal revenues and costs so they can in turn maximize their profits.

Here is a video showing how calculus can be used to determine marginal revenue and cost:

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| ''' Economics: Marginal Cost and Revenue'''
|-
|style="background: padding:12px;"|
<center>This video shows how calculus can be used to determine marginal revenue and cost </center>

{{#ev:youtube | 52ANs_PZQjI| 400}}
|}

Integrals are also predominant in economics, especially in the areas of consumer and producer surplus and deadweight loss. Calculus is also applied in the topics of elasticity, for example when given a formula to find elasticity rather then a numerical value, the derivative of that formula must be found in order to do so.
In its most basic form, the logic and deductive knowledge theorized in calculus is transcended through the application of economics. Logic and problem solving are prevalent in both micro and macro economics in analyzing how market forces work, externalities, growth rates, elasticity, and etc. The techniques and skills I gain though my calculus courses will definitely be a great asset to my studies of economics.

File:Economics2.jpg

2011-01-28T07:59:28Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 12

2011-01-28T07:30:00Z

SabrinaPannu:

==Team Valais: Homework 12==

==='''Problem'''===

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Problem
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

|}
==='''Solution'''===

In order to solve this problem we first have to look at the general shape of the graph. When graphing the function it will look very similar to the following graph:
[[File:Wolframgraph.png]]

For the first problem to solve we have to put the k-value after the whole equation in order to move the horizontal asymptote to any direction.
The formula would look like :

Now if we change this number to k=2 our horizontal asymptote would be at k+1 (3):
[[File:Graph2.png]]

So as we can see our horizontal asymptote is at 3 when k=2.

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 12

2011-01-28T07:19:33Z

SabrinaPannu:

{| class="wikitable" width="800px"
!style="background: #777777; color: #D0D0D0; text-align: left; padding:3px;" width="100%"|Team Problem
|-
|style="background: #FFFFFF; padding:12px;"|
Starting with the function
:<math>P(t) = \frac{1}{1+e^{-t}}</math>
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:
* Change the height of the horizontal asymptote on the right, we'll denote it by <math>K</math>.
* Change the <math>y</math>-intercept to any number between <math>0</math> and <math>K</math>
BONUS (just the point below, not what comes after)
* Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear).

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity).

When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.

This should all be done on a dedicated page of the wiki whose address should look like
* wiki.ubc.ca/Course:MATH110/003/Teams/YOURTEAMNAME/Homework_12
* wiki.ubc.ca/Course:MATH110/003/Teams/YOURTEAMNAME/Homework/Homework_12
|}

In order to solve this problem we first have to look at the general shape of the graph. When graphing the function it will look very similar to the following graph:
[[File:Wolframgraph.png]]

For the first problem to solve we have to put the k-value after the whole equation in order to move the horizontal asymptote to any direction.
The formula would look like :

Now if we change this number to k=2 our horizontal asymptote would be at k+1 (3):
[[File:Graph2.png]]

So as we can see our horizontal asymptote is at 3 when k=2.

File:Graph2.png

2011-01-28T07:15:34Z

SabrinaPannu:

File:Wolframgraph.png

2011-01-28T07:14:22Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 12

2011-01-28T07:13:19Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Teams/Valais

2011-01-26T00:25:12Z

SabrinaPannu:

{{
Infobox MATH110 Teams
| team name = Valais
| member 1 = Albert König
| member 2 = Charlene Ng
| member 3 = Jonathan Rothwell
| member 4 = [[User:SabrinaPannu|Sabrina Pannu]]
}}
In workshop G.
==Contact Information==
* Sabrina Pannu : # 604-786-1757 ''email:'' pannu7@msn.com
'''Everyone please add your contact information!''' [[User:SabrinaPannu|SabrinaPannu]]

==Homework==
[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_11 Homework 11]

[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_12 Homework 12]
==Key Words==
* Sabrina - sine

* Jonathan - tangent

* Albert - cotangent

* Charlene - cosine

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 12

2011-01-26T00:17:55Z

SabrinaPannu: Created page with "h12"

h12

Course:MATH110/Archive/2010-2011/003/Teams/Valais

2011-01-24T16:10:46Z

SabrinaPannu: /* Key Words */

{{
Infobox MATH110 Teams
| team name = Valais
| member 1 = Albert König
| member 2 = Charlene Ng
| member 3 = Jonathan Rothwell
| member 4 = Sabrina Pannu
}}
In workshop G.
==Contact Information==
* Sabrina Pannu : # 604-786-1757 ''email:'' pannu7@msn.com ''skype:'' sabrina.pannu ''facebook:'' Sabrina Pannu
'''Everyone please add your contact information!''' [[User:SabrinaPannu|SabrinaPannu]]

==Homework==
[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_11 Homework 11]

==Key Words==
* Sabrina - sine

* Jonathan - tangent

* Albert - cotangent

* Charlene -cosine

Course:MATH110/Archive/2010-2011/003/Teams/Valais

2011-01-24T16:10:01Z

SabrinaPannu:

{{
Infobox MATH110 Teams
| team name = Valais
| member 1 = Albert König
| member 2 = Charlene Ng
| member 3 = Jonathan Rothwell
| member 4 = Sabrina Pannu
}}
In workshop G.
==Contact Information==
* Sabrina Pannu : # 604-786-1757 ''email:'' pannu7@msn.com ''skype:'' sabrina.pannu ''facebook:'' Sabrina Pannu
'''Everyone please add your contact information!''' [[User:SabrinaPannu|SabrinaPannu]]

==Homework==
[http://wiki.ubc.ca/Course:MATH110/003/Teams/Valais/Homework_11 Homework 11]

==Key Words==
Sabrina - sine
Jonathan - tangent
Albert - cotangent
Charlene -cosine

Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 11

2011-01-19T03:45:35Z

SabrinaPannu:

team valais homework 11

1) The model that we are going to use in this part is:
[[File:math011.png]]

The above models based on the information given. At first we are told that we have to spend 100$ for the first 20 units bought. Later on we are told that foe any extra unit we have to pay 7$. The variable x in the above equation indicates the amount of units bought minus the initial 20units that have to be subtracted because they are worth 100$.
2) Using the above equation our model predicts the following value for 150 units:
[[File:math022.png]]

What we can conclude from the above equation is that for 150 units we will have a Total cost of 1010$.
3) Now in order to find the Average Cost of per item produced we have to do some modification to the model introduced in no.1. We have to divide the whole equation by ‘x’ which is the total amount of units bought:

Now if we imply this formula for 150 units bought, we will have the following calculation:
[[File:math033.png]]

This means our average cost for 150 units is 6.73$ which is below our Marginal Cost
Now if we increase number of units bought to 250, we will have the following calculation:

[[File:math044.png]]

This indicates that for buying 250 units we will have to pay an average of 6.84$, again below the Marginal cost.
1) A possible example of having the average cost to remain constant is to use the following equation:
[[File:math00.png]]

Our model: C(x)= 100 + 7(x-20), x>20
It is a linear equation. 100 is the fixed cost for the production of 20 units and 7 is the variable cost. (x-10) is for the number of units to be multiplied by 7 to get the addition cost of producing 1 more unit.

C(150) = 100 + 7(150-20)= 1010

Average Cost is the derivative of C(x). A(x) = 40x^(-2)
The average cost per item increases as production levels increase.

Other models (not necessarily linear) for which you get other behaviours such as:
1. The average cost remains constant as production increases.
A(x) = 5
C(x) = 5x

2. The average cost diminishes as production increases.
A(x) = 1/x^2
C(x) = 1/x

3. The average cost increases as production increases.
A(x) = x
C(x) = x^2

4.You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

A(x) = (x-10)^2 + 10
C(x) = x(x-10)^2 + 100

You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.

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Course:MATH110/Archive/2010-2011/003/Teams/Valais

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Course:MATH110/Archive/2010-2011/003/Teams/Valais

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Course:MATH110/Archive/2010-2011/003/Teams/Valais/Homework 11

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Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-06T08:28:16Z

SabrinaPannu: /* Tutorial Videos */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}
----

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

----

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| ''' Textbook Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
''Problems from the Just-In-Time Textbook:''

'''Section 4.2. Lines and Their Equations

Page 60. #11-18
'''

|}

----

====Helpful Links====
----
Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===
----
[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====
----
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

'''3) What is the distance between the points ( 1, -3 ) and ( 0, -5) ?'''' (look on p.58 in Just-in-time for a diagram)

<center>Answers:
1)6.08

2)38.01 

3)2.24
|}

----

====Helpful Links====
----
http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}

----
====Examples====

----

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Example Question 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}
----

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===
----
<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''1.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>

|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''2.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

|}
----
====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}
----

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}
----

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]
----
===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!
----

====Examples====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Examples
|-
|style="background: #FBBBB9; padding:12px;"|

<center>
*'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b.

Simplifying this, we get 5=-2+b-->7=b.

Therefore, the answer is y=-1x+7</center>

<center>
*'''Find the equation of a line given the point (3,4) and slope 2/3.'''
<center> We know that m=2/3 and x=3 and y=4 so we can plug into the slope intercept formula (y=mx+b), resulting in 4=2/3(4)+b.

Simplifying this, we get 4= 8/3+b--> b=4/3.

Therefore, the answer is y=2/3x+4/3

<center>
*'''Find the equation of a line given the point (-4,9) and slope 0.'''
<center> We know that m=0 and x=-4 and y=9 so we can plug into the slope intercept formula (y=mx+b), resulting in 9=0(-4)+b.

Simplifying this, we get 4= 0+b--> b=4.

Therefore, the answer is y=4<center>
|}

------

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video- Equation of a Line Using a Point and the Slope (Method 1) '''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | 5lp3edqRalg| 400}}
|}

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video-Equation Of a line with a Point and the Slope (Method 2) '''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | jBs-79C3wwM| 400}}
|}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}
----
====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>
----

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-06T07:25:33Z

SabrinaPannu: /* Practice Problems */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}
----

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

----

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| ''' Textbook Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
''Problems from the Just-In-Time Textbook:''

'''Section 4.2. Lines and Their Equations

Page 60. #11-18
'''

|}

----

====Helpful Links====
----
Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===
----
[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====
----
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

'''3) What is the distance between the points ( 1, -3 ) and ( 0, -5) ?'''' (look on p.58 in Just-in-time for a diagram)

<center>Answers:
1)6.08

2)38.01 

3)2.24
|}

----

====Helpful Links====
----
http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}

----
====Examples====

----

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Example Question 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}
----

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===
----
<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''1.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>

|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''2.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

|}
----
====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}
----

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}
----

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]
----
===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!
----

====Examples====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Examples
|-
|style="background: #FBBBB9; padding:12px;"|

<center>
*'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b.

Simplifying this, we get 5=-2+b-->7=b.

Therefore, the answer is y=-1x+7</center>

<center>
*'''Find the equation of a line given the point (3,4) and slope 2/3.'''
<center> We know that m=2/3 and x=3 and y=4 so we can plug into the slope intercept formula (y=mx+b), resulting in 4=2/3(4)+b.

Simplifying this, we get 4= 8/3+b--> b=4/3.

Therefore, the answer is y=2/3x+4/3

<center>
*'''Find the equation of a line given the point (-4,9) and slope 0.'''
<center> We know that m=0 and x=-4 and y=9 so we can plug into the slope intercept formula (y=mx+b), resulting in 9=0(-4)+b.

Simplifying this, we get 4= 0+b--> b=4.

Therefore, the answer is y=4<center>
|}

------

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}
----
====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>
----

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-06T07:18:12Z

SabrinaPannu: /* Examples */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}
----

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

----

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====
----
Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===
----
[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====
----
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

'''3) What is the distance between the points ( 1, -3 ) and ( 0, -5) ?'''' (look on p.58 in Just-in-time for a diagram)

<center>Answers:
1)6.08

2)38.01 

3)2.24
|}

----

====Helpful Links====
----
http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}

----
====Examples====

----

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Example Question 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}
----

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===
----
<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''1.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>

|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''2.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

|}
----
====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}
----

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}
----

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]
----
===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!
----

====Examples====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Examples
|-
|style="background: #FBBBB9; padding:12px;"|

<center>
*'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b.

Simplifying this, we get 5=-2+b-->7=b.

Therefore, the answer is y=-1x+7</center>

<center>
*'''Find the equation of a line given the point (3,4) and slope 2/3.'''
<center> We know that m=2/3 and x=3 and y=4 so we can plug into the slope intercept formula (y=mx+b), resulting in 4=2/3(4)+b.

Simplifying this, we get 4= 8/3+b--> b=4/3.

Therefore, the answer is y=2/3x+4/3

<center>
*'''Find the equation of a line given the point (-4,9) and slope 0.'''
<center> We know that m=0 and x=-4 and y=9 so we can plug into the slope intercept formula (y=mx+b), resulting in 9=0(-4)+b.

Simplifying this, we get 4= 0+b--> b=4.

Therefore, the answer is y=4<center>
|}

------

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}
----
====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>
----

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-06T04:39:41Z

SabrinaPannu:

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}
----

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

----

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====
----
Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===
----
[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====
----
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

'''3) What is the distance between the points ( 1, -3 ) and ( 0, -5) ?'''' (look on p.58 in Just-in-time for a diagram)

<center>Answers:
1)6.08

2)38.01 

3)2.24
|}

----

====Helpful Links====
----
http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}

----
====Examples====

----

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Example Question 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}
----

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===
----
<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''1.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>

|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"|
'''2.) What is the equation of this line?'''
|style="background: ; padding:12px;"|

<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

|}
----
====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}
----

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}
----

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]
----
===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!
----

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

'''Find the equation of a line given the point (3,4) and slope 2/3.'''
<center> we know that m=2/3 and x=3 and y=4 so we can plug into the slope intercept formula (y=mx+b), resulting in 4=2/3(4)+b. Simplifying this, we get 4= 8/3+b--> b=4/3. Therefore, the answer is y=2/3x+4/3<center>

'''Find the equation of a line given the point (-4,9) and slope 0.'''
<center> we know that m=0 and x=-4 and y=9 so we can plug into the slope intercept formula (y=mx+b), resulting in 9=0(-4)+b. Simplifying this, we get 4= 0+b--> b=4. Therefore, the answer is y=4<center>

------

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}
----
====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>
----

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:26:04Z

SabrinaPannu: /* Example */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Examples====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Example Question 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:24:26Z

SabrinaPannu:

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Example====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| Tips and Tricks
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:22:41Z

SabrinaPannu:

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Example====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|}

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:22:00Z

SabrinaPannu: /* Tips and Tricks */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Example====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

* When given a graph of a line and asked to find the equation, the slope intercept form ( y=mx+b )is generally the easiest to use because the y-intercept b can be easily identified from the graph.

* The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

* When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

* When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

|{

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:18:26Z

SabrinaPannu: /* Example */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Example====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----
When given a graph of a line and asked to find the equation, the slope intercept form (y=mx+b)is generally the easiest to use because the y-intercept b can be easily identified from the graph.
The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-05T08:17:54Z

SabrinaPannu:

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What Does It Mean For Two Lines To Be Parallel And/Or Perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Parallel Lines Example 2'''
|-
|style="background: #ADDFFF; padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 2'''
|-
|style="background: #E6A9EC;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

----

===How to Compute the Distance between Two Points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

{| class="wikitable"
!style="background: #6960EC; text-align: left; padding:3px;"| '''Example 1'''
|-
|style="background: #ADDFFF; padding:12px;"|

[[File:basicskills-3.jpg]]
|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points ( -2, 7 ) and ( 4, 6 ) ?''''

'''2) What is the distance between the points ( 5, 6 ) and ( -12, 40 ) ?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

----

===How To Compute The Equation of a Line Given Two Points===

----
{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WHow To Compute The Equation of a Line Given Two Points
|-
|style="background: #FBBBB9; padding:12px;"|

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

*'''Determine the slope of the line'''.

''Recall that the slope is: the rise over run of a line.''

The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line).
[[File:slope.jpg]]

This means that you find the numerical value of the slope of any line by [[File:slope1.GIF]]
Let us now call this number (the slope) ''m.''

*'''Next, you simply plug in the values from one of the line's points and the slope into the point slope equation:''' <math>y - y_1 = m( x - x_1 )\,</math> Either point can be used as long as the x-coordinate and y-coordinate are from the same point.

If you want to put this into slope intercept form <math>y = mx + b,\,</math>, you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
|}
====Example====

{| class="wikitable"
!style="background: #CA226B; text-align: left; padding:3px;"| '''Perpendicular Lines Example 1'''
|-
|style="background: #E6A9EC;padding:12px;"|

Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

1* <math>(-4-1))/(4-2)= -5/2,\,</math>.

SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

2* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>

Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

|}
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====

http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php
----

===What Is The Equation Of A Line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>

<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' ''(where k represents any real number that is the value
of the x-coordinate of the graph).''''

====Examples====
----
'''What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----
When given a graph of a line and asked to find the equation, the slope intercept form (y=mx+b)is generally the easiest to use because the y-intercept b can be easily identified from the graph.
The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How To Compute The Equation Of A Line Given Its Slope And A Point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

File:Slope1.GIF

2010-12-05T07:58:58Z

SabrinaPannu:

File:Slope.jpg

2010-12-05T07:55:35Z

SabrinaPannu:

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-03T19:06:29Z

SabrinaPannu:

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What does it mean for two lines to be parallel and/or perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Parallel Lines Example Question 1'''
|-
|style="background: #FFE6EA; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''Parallel Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #3BB9FF text-align: left; padding:3px;"| '''Perpendicular Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Perpendicular Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

===How to compute the distance between two points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

[[File:basicskills-3.jpg]]
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points (-2,7) and (4,6)?''''

'''2) What is the distance between the points(5,6) and (-12,40)?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

===How to compute the equation of a line given two points===

----

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

1*'''Determine the slope of the line'''. Recall that the slope is: the rise over run of a line. The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line). This means that you find the numerical value of the slope of any line by <math>[rise:(y_2-y_1)/run:(x_2-x_1)]</math> Let us now call this number ''m.''

2*Next, you simply '''plug in the values''' from one of the line's points and the slope into the most basic formula for a line: <math>y - y1 = m( x - x1 </math> For y1 and x1, either point can be used as long as the x-coordinate and y-coordinate are from the same point. If you want to put this into slope intercept form <math>y = mx + b,\,</math> you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
====Example====
----
Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

1* <math>(-4-1))/(4-2)= -5/2,\,</math>. SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

2* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>
Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php

===What is the equation of a line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>
<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' (where k represents any real number that is the value
of the x-coordinate of the graph).''

====Examples====
----
'''1)What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----
When given a graph of a line and asked to find the equation, the slope intercept form (y=mx+b)is generally the easiest to use because the y-intercept b can be easily identified from the graph.
The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How to compute the equation of a line given its slope and a point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
|}
----

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
|}

====Helpful Links====
----
<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>

Course:MATH110/Archive/2010-2011/003/Groups/Group 02/Basic Skills Project

2010-12-03T18:57:27Z

SabrinaPannu: /* Examples */

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
|-
|style="background: #FBBBB9; padding:12px;"|
Welcome to [http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_02 Group 2]`s page. Our contribution to the Basic Skills Project is the topic of '''Distance and Lines.''' On this page you will find
* detailed step-by-step examples
* tutorial videos
* tips and tricks
* practice problems
* helpful links
on the all of the sub-topics.

[[File:Index.jpg]]Please Visit [http://www.youtube.com/user/math110group2 Group 2's Youtube Page] for
[http://www.youtube.com/user/math110group2#p/u videos created by us] and [http://www.youtube.com/user/math110group2#p/f videos we find informative and helpful.]

|}

===What does it mean for two lines to be parallel and/or perpendicular?===
----

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| How do you know if two lines are parallel?
|style="background: #FFE6EA; padding:12px;"|
<center>'''Their slopes are the same!'''</center>
|}

[[File:parallelpic.gif]]

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| How do you know if two lines are perpendicular?
|style="background: #FBBBB9; padding:12px;"|
<center>'''When you multiply their slopes, you get -1!'''</center>

|}

[[File:Per.gif]]
====Examples====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Parallel Lines Example Question 1'''
|-
|style="background: #FFE6EA; padding:12px;"|

'''
'''Problem: Determine whether the graphs of y = -3x + 5 and 4y = -12x + 20 are parallel lines.'''''
*Solve for y for both graphs
y = -3x + 5 >already solved.

4y = -12x + 20 >Solve for y

4y=-12x+20 >Divide both sides by 4
to get:

y = -3x + 5

''''''The slope-intercept equations are the same. The two equations have the same graph and the same slope; thus, they are parallel.''''''
|}

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Parallel Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''''Find the equation of the line that is: parallel to y = 2x + 1 and passes though the point (5,4)'''''

''Recall that parallel lines have the same slope!''

1.) The first step is to find the slope of <math>y = 2x + 1.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is 2.

The slope of <math> y=2x+1 is : 2 </math>

2.) The next step is to substitute the slope 2 into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

We obtain: <math> y - y_1 = 2(x - x_1)</math>

And now we must put in the point (5,4):

<math>y - 4 = 2(x - 5)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math>y - 4 = 2x - 10</math> >solve for y

<math>y = 2x - 6 </math>'''>answer'''

'''The lines have both the same slope: [2] making them parallel.'''
|}

{| class="wikitable"
!style="background: #3BB9FF text-align: left; padding:3px;"| '''Perpendicular Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''Determine whether the lines 5y = 4x + 10 and 4y = -5x + 4 are perpendicular.
'''

Find the slope-intercept equations for both lines
by solving for y.

y = (4/5)x + 2

y = -(5/4)x + 1

(4/5) MULTIPLIED BY -(5/4) = '''-1'''

'''The product of the slopes is -1, so the lines are perpendicular.'''
|}

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Perpendicular Lines Example Question 2'''
|-
|style="background: #FFE6EA;padding:12px;"|
Example:
'''
'''Find the equation of the line that is perpendicular to y = -4x + 10 and passes though the point (7,2)'''''

1.)The first step is to find the slope of <math>y= -4x + 10.</math>
Recall <math>y = mx + b\,</math>
with m being slope. Therefore, the slope of this line is -4

'''>'''The slope of<math> y=-4x+10 is: -4</math>

The negative reciprocal of that slope is:
<math> m=\frac{1}{-4}=\frac{1}{4}</math>

So the perpendicular line will have a slope of 1/4.

NOTE: For more information on negative recipricals, see tips and tricks.

2.) The next step is to substitute the slope (1/4) into the point slope equation
of a line which is <math>y - y_1 = m( x - x_1 )\,</math>

<math> y - y1 = (1/4)(x - x1)</math>

And now put in the point (7,2):

<math> y - 2 = (1/4)(x - 7)</math> '''>answer'''

This is the answer, however we can also put this answer into slope-intecept form or
<math>y = mx + b\,</math> form:

<math> y - 2 = x/4 - 7/4</math> >solve for y

<math>y = x/4 + 1/4 </math> '''>answer'''

'''When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular'''

|}

----

====Tips and Tricks====

{| class="wikitable"
!style="background: #F75D59; text-align: left; padding:3px;"| '''TIP: Know Your Negative Reciprocals'''
|-
|style="background: #FBBBB9;padding:12px;"|

*Knowing how to find a negative reciprocal is a useful skill for it helps you find a perpendicular line to an equation. (See Perpindicular Lines Example Question Above)

For example:
For an equation with a slope of 5x, the recriprocal would be: <math>\frac{-1}{5}</math> x

How do we get <math>\frac{-1}{5}</math> X

It`s quite simple:
*Take 5x (which can also be written as <math>\frac{5}{1}</math> X and flip the bottom and the top, so it becomes: <math>\frac{1}{5}</math> X
This is the RECIPRICOAL, but we want the NEGATIVE RECIPROCAL so we must do the next step:
*Change the positive sign to a negative sign so it becomes <math>\frac{-1}{5}</math> X

'''TIP:''' It is important to remember the difference between simply a recipricoal and a negative reciprocoal as they can be easily confused.
For practice see practice problems below.

|}

----

====Tutorial Videos====

{{#ev:youtube | Rew54K6mYUo| 400}}
{{#ev:youtube | oZg7O-3GLNI| 400}}

----

====Practice Problems====

{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Negative Reciprocal Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''What is the negative reciprocal of the following:'''

'''1.) 5'''

'''2.) 4/9'''''

'''3.) -7/3'''

Answers:

1.) -1/5
2.) -9/4
3.) 3/7


|}


----

====Helpful Links====

Here are some websites that have a great amount of information:

[http://www.purplemath.com/modules/slope2.html Purple Math] - A great website with lots of examples.

[http://www.beaconlearningcenter.com/documents/1750_01.pdf] - This fantastic PDF has lots of practice problems and tons of examples.

===How to compute the distance between two points===

[[File:basicskills-1.jpg]]
[[File:basicskills-2.jpg]]

----
====Examples====

[[File:basicskills-3.jpg]]
----
====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | KP5xEzoABic| 400}}

|}

----

====Practice Problems====
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

'''1) What is the distance between the points (-2,7) and (4,6)?''''

'''2) What is the distance between the points(5,6) and (-12,40)?''''

<center>Answers:
1)6.08

2)38.01 
|}

----
====Helpful Links====

http://www.tpub.com/math2/2.htm

http://www.purplemath.com/modules/distform.htm

===How to compute the equation of a line given two points===

----

If you have two points on a line you can construct the general equation for the that line. It is achieved easily by breaking it down into two simple tasks:

1*'''Determine the slope of the line'''. Recall that the slope is: the rise over run of a line. The slope of the line is just the change in Y (values of the y-coordinates of the points on the line) divided by the change in X (values of the x-coordinates of the points on the line). This means that you find the numerical value of the slope of any line by <math>[rise:(y_2-y_1)/run:(x_2-x_1)]</math> Let us now call this number ''m.''

2*Next, you simply '''plug in the values''' from one of the line's points and the slope into the most basic formula for a line: <math>y - y1 = m( x - x1 </math> For y1 and x1, either point can be used as long as the x-coordinate and y-coordinate are from the same point. If you want to put this into slope intercept form <math>y = mx + b,\,</math> you can plug in the y-coordinate and x-coordinate from either point as well as m (slope) into the equation to find b (the y-intercept).
====Example====
----
Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above.

1* <math>(-4-1))/(4-2)= -5/2,\,</math>. SO, <math>m = -5/2,\,</math> or <math>-2.5,\,</math>

2* Using the first point: <math>y-1= -2.5(x-2),\,</math> or <math>y = -2.5x + 6,\,</math>
Using the second point: <math>y-(-4) = -5/2(x-4)\,</math> or <math>y = -5/2x + 6\,</math>

====Tutorial Videos====
----

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
|-
|style="background: padding:12px;"|
<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | hsc9POhVPh8| 400}}

|}
----

====Practice Problems====

----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|

<center>Here are a few sets of points that you can find equations for if you want practice. If you want the answers they can be sent by email upon request:</center>

''''1. (4, 3) and (6, 2)''''

''''2. (-8, -2) and (13, 9)''''

''''3. (2.6, 1) and (-pi, 11)''''
|}

====Helpful Links====
----
http://www.mathsisfun.com/algebra/line-equation-2points.html

http://www.tutorvista.com/content/math/geometry/straightlines/two-point-form.php

===What is the equation of a line?===

<center>'''General equation of a straight line'''</center>
<center>'''<math>Ax+By=C\,</math>'''</center>
<center>''where a,b & c are constants and a & b cannot both be zero.''</center>
<center>'''Slope-intercept form of a line'''</center>
<center>'''<math>y = mx + b\,</math> '''</center>
<center>''where m is the slope of the line and b is the y-intercept of the graph of the line.'' </center>
<center>'''Point-slope form'''</center>
<center>'''<math>y - y_1 = m( x - x_1 )\,</math> )''' </center>
<center>''for a line through a point with coordinates (x1, y1) and slope m.''</center>

<center>'''Horizontal lines'''</center>
<center>[[File:horizontal line 3.gif]]</center>
<center>''As shown in the graph above, every point on a horizontal line has the same y-coordinate so
the equation of a horizontal line is '''y=k''' (where k represents any real number that is the value
of the y-coordinate of the graph).''
<center>'''Vertical lines'''</center>
<center>[[File:vertical line.gif]]</center>
<center>'' As shown in the graph above, every point on a vertical line has the same x-coordinate so
the equation of a vertical line is '''x=k''' (where k represents any real number that is the value
of the x-coordinate of the graph).''

====Examples====
----
'''1)What is the equation of this line?'''
<center>[[File:Example.gif]]</center>
<center>The graph crosses the y-axis at zero, so the easiest equation to use for this graph is the slope intercept form. We know that b=0, so we need to find the slope m. Using the point (1,2) and the point (2,4)from the graph, we can find the slope with the equation y2-y1/x2-x1. 4-2/2-1 = 2/1 so the slope is 2. Plugging that into the slope intercept form for m,the answer is '''y=2x.'''</center>
'''2) What is the equation of this line? '''
<center>[[File:example 2.gif]]
<center> The graph shows us the y-intercept, so the slope intercept form is the easiest equation to use. We know that b=1, so we just need to find the slope m. Using the point (2,5) and the point (0,1) from the graph, we can find the slope with the equation y2-y1/x2-x1. 1-5/0-2 =-4/-2 =2 so the slope is 2. Plugging that into the slope intercept form for m, the answer is '''y=2x+1.'''</center>

====Tips and Tricks====
----
When given a graph of a line and asked to find the equation, the slope intercept form (y=mx+b)is generally the easiest to use because the y-intercept b can be easily identified from the graph.
The slope m can usually also be easily calculated by using two points from the graph and the slope formula y2-y-1/x2-x1.

When given a point and its slope, the slope intercept form is also generally the easiest because the slope and y and x can be plugged into the equation to find b.

When given the coordinates of two points and asked to find the equation, the point-slope form (y-y1=m(x-x1) is usually the easiest to use because the slope can easily be calculated by using the slope formula. After finding the formula, plug in the slope for m in the point-slope formula and x1 and y1 from either given point, as long as both coordinates are from the same point.

====Tutorial Videos====
----

{{#ev:youtube | hh6RAEPlza4| 400}}

{{#ev:youtube | GFM8NOe_XM4| 400}}
{{#ev:youtube | rNQ36DK6aBk| 400}}

====Practice Problems====
----
{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
|-
|style="background: #FFE6EA;padding:12px;"|
'''
'''1) Determine an equation for a line through the points (5,5) and (4,-1)'''
'''

'''2) Determine an equation for a line for the point (2,6) with slope -3.
'''

'''3) Determine an equation for a line that passes through the point (1,1) with slope 2.'''

Answers:

'''1)'''y-5=6(x-5)

'''2)'''y=-3x+12

'''3)'''y-1=2(x-1)

<center> '''Practice Questions from Just-In-Time textbook'''</center>
'''Page 60 #9, #11, #13'''

|}

====Helpful Links====
----

1)[http://www.mathsisfun.com/equation_of_line.html Equation of straight line]
2)[http://www.purplemath.com/modules/strtlneq.htm Slope Intercept Form]
3)[http://www.purplemath.com/modules/strtlneq2.htm Point Slope Form]

===How to compute the equation of a line given its slope and a point.===

-----

[[File:Cartoon.math.gif]]

The equation of a line is defined by the equation
<math>y = mx + b,\,</math>

All these variables represent some specific and important part of a curve that define that shape of it:

''m = slope

b = y intercept''

If the slope is given, then the only thing that has to be done is that it must be plugged it into the equation: <math>y = mx + b,\,</math>

Now that you have a value of m, the next thing you do is plug in the values for x and y.

This means that the point given as the x value is plugged into the equation as x. The same is done with the y point into the equation.

When this is done the only thing needed is solve for b.

Now that you have m and b, you plug those values back into the original equation (<math>y = mx + b\,</math> ).

And VOILA! You have the equation for the curve!

====Examples====
----
'''Find the equation of a line given the point (2,5) and slope -1.'''
<center>We know that m = -1 and x=2 and y=5 so we can plug those into the slope intercept formula (y=mx+b), resulting in 5=-1(2)+b. Simplifying this, we get 5=-2+b-->7=b. Therefore, the answer is y=-1x+7</center>

====Tutorial Videos====

{| class="wikitable"
!style="background: #FBBBB9; text-align: left; padding:3px;"| '''Original Tutorial Video'''
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<center>'''YOUTUBE CHANNEL: math110group2'''''
http://www.youtube.com/user/math110group2'''</center>

{{#ev:youtube | WKAUmRUaai8| 400}}
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====Practice Problems====
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{| class="wikitable"
!style="background: #3BB9FF; text-align: left; padding:3px;"| '''Practice Problems'''
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'''1) Find the equation of a line through the point (-1,3) with slope 2.'''

'''2) Find the equation of a line through (-2,-10) with slope 4.'''
<center>Answers: 1)y=2x+5 2)y=4x-2
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====Helpful Links====
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<center>http://www.nipissingu.ca/calculus/tutorials/linear.html
</center>
<center> http://www.tpub.com/math2/6.htm </center>