Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 12

MATH152 April 2016

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Question A 12
Find the distance from the point (1, 2) to the line $x+y=0$ in the plane.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The distance between a point to a line is the length of the line segment which joins the point to the line and is perpendicular to the line.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Denote the given line $y+x=0$ by $L_{1}$. We first need to find a line $L_{2}$ which passes the point (1 , 2) and is normal to the line $L_{1}:y+x=0$. The distance from the point (1 , 2) to the line $L_{1}$, is exactly the distance between (1,2) and intersection points of these two lines $L_{1}$ and $L_{2}$. As the slope of the line $y+x=0$ is -1, the slope of the normal line $L_{2}$ should be 1. Therefore, the normal line passing through (1, 2) is $y-2=(x-1)$ ( i.e $y=x+1$). The intersection of $y+x=0$ , and $y=x+1$ is $(-1/2,1/2)$ . Thus the distance between (1, 2) and ( -1/2, 1/2) is ${\sqrt {(1+1/2)^{2}+(2-1/2)^{2}}}=\color {blue}{\frac {3{\sqrt {2}}}{2}}$.