How do you graph the line that passes through (-1,5) perpendicular to the graph 5x−3y−3=0?

1 Answer
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Apr 12, 2018

y=−35x+225 graph{-3/5x+22/5 [-10, 10, -5, 5]} #

Explanation:

First, get the equation into the form y=mx+c
3y=5x−3
y=53x−1

The gradient of the perpendicular line is the negative reciprocal of the original line. The gradient of the original line is 53, so the gradient of the perpendicular line is −35

Put this into the equation y=mx+c

y=−35x+c

To find c, plug in values (given by the coordinates in the question) and solve

5=−35(−1)+c
5=35+c
c=225

The equation of the line is y=−35x+225

Now for graphing.
You know the line passes through the point (−1,5). Plot this point.
You know that the y-intercept is (0,225). Plot this point.
The gradient of the line is −35, meaning that for every 3 down you go, you go 5 to the right. Starting from either of the points you've already plotted, go 3 down and 5 to the right. Plot this point.
Now you have 3 points, join them together and extend the line.

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