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

WELCOME TO GROUP TWO'S PAGE: DISTANCE AND LINES
Welcome to 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. Please Visit Group 2's Youtube Page for videos created by us and videos we find informative and helpful.

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

How do you know if two lines are parallel?	Their slopes are the same!

How do you know if two lines are perpendicular?	When you multiply their slopes, you get -1!

Examples

Parallel Lines Example 1
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.'

Parallel Lines Example 2
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 $y=2x+1.$ Recall $y=mx+b\,$ with m being slope. Therefore, the slope of this line is 2. The slope of $y=2x+1is:2$ 2.) The next step is to substitute the slope 2 into the point slope equation of a line which is $y-y_{1}=m(x-x_{1})\,$ We obtain: $y-y_{1}=2(x-x_{1})$ And now we must put in the point (5,4): $y-4=2(x-5)$ >answer This is the answer, however we can also put this answer into slope-intecept form or $y=mx+b\,$ form: $y-4=2x-10$ >solve for y $y=2x-6$ >answer The lines have both the same slope: [2] making them parallel.

Perpendicular Lines Example 1
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.

Perpendicular Lines Example 2
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 $y=-4x+10.$ Recall $y=mx+b\,$ with m being slope. Therefore, the slope of this line is -4 >The slope of $y=-4x+10is:-4$ The negative reciprocal of that slope is: $m={\frac {1}{-4}}={\frac {1}{4}}$ 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 $y-y_{1}=m(x-x_{1})\,$ $y-y1=(1/4)(x-x1)$ And now put in the point (7,2): $y-2=(1/4)(x-7)$ >answer This is the answer, however we can also put this answer into slope-intecept form or $y=mx+b\,$ form: $y-2=x/4-7/4$ >solve for y $y=x/4+1/4$ >answer When you multiply the slopes of the lines [-4] and [1/4] you get -1, making the lines perpendicular

Tips and Tricks

TIP: Know Your Negative Reciprocals
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: ${\frac {-1}{5}}$ x How do we get ${\frac {-1}{5}}$ X It`s quite simple: Take 5x (which can also be written as ${\frac {5}{1}}$ X and flip the bottom and the top, so it becomes: ${\frac {1}{5}}$ 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 ${\frac {-1}{5}}$ 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

Practice Problems

Negative Reciprocal Practice Problems
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

Textbook Practice Problems
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:

Purple Math - A great website with lots of examples.

[1] - This fantastic PDF has lots of practice problems and tons of examples.

How to Compute the Distance between Two Points

Examples

Example 1

Tutorial Videos

Original Tutorial Video
YOUTUBE CHANNEL: math110group2 http://www.youtube.com/user/math110group2

Practice Problems

Practice Problems
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) _{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

WHow 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: 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 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: $y-y_{1}=m(x-x_{1})\,$ 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 $y=mx+b,\,$ , 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

Example Question 1
Let us choose two random points: (2, 1) and (4, -4). Now we will perform the steps outlined above. $(-4-1))/(4-2)=-5/2,\,$ . SO, $m=-5/2,\,$ or $-2.5,\,$ Using the first point: $y-1=-2.5(x-2),\,$ or $y=-2.5x+6,\,$ Using the second point: $y-(-4)=-5/2(x-4)\,$ or $y=-5/2x+6\,$

Tutorial Videos

Original Tutorial Video
YOUTUBE CHANNEL: math110group2 http://www.youtube.com/user/math110group2

Practice Problems

Practice Problems
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: '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?

General equation of a straight line $Ax+By=C\,$ where a,b & c are constants and a & b cannot both be zero. Slope-intercept form of a line $y=mx+b\,$ where m is the slope of the line and b is the y-intercept of the graph of the line. Point-slope form $y-y_{1}=m(x-x_{1})\,$ ) for a line through a point with coordinates (x₁, y₁) and slope m.

Horizontal lines

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).

Vertical lines

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?	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.

2.) What is the equation of this line?	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.

Tips and Tricks

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

Practice Problems

Practice Problems
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) Practice Questions from Just-In-Time textbook Page 60 #9, #11, #13

Helpful Links

1)Equation of straight line 2)Slope Intercept Form 3)Point Slope Form

How To Compute The Equation Of A Line Given Its Slope And A Point.

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The equation of a line is defined by the equation $y=mx+b,\,$

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: $y=mx+b,\,$

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 ( $y=mx+b\,$ ).

And VOILA! You have the equation for the curve!

Examples

Examples
Find the equation of a line given the point (2,5) and slope -1. 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 Find the equation of a line given the point (3,4) and slope 2/3. 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 Find the equation of a line given the point (-4,9) and slope 0. 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

Tutorial Videos

Original Tutorial Video- Equation of a Line Using a Point and the Slope (Method 1)
YOUTUBE CHANNEL: math110group2 http://www.youtube.com/user/math110group2

Original Tutorial Video-Equation Of a line with a Point and the Slope (Method 2)
YOUTUBE CHANNEL: math110group2 http://www.youtube.com/user/math110group2

Practice Problems

Practice Problems
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. _{Answers: 1)y=2x+5 2)y=4x-2}

Helpful Links

http://www.nipissingu.ca/calculus/tutorials/linear.html http://www.tpub.com/math2/6.htm