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

1 Answer
Apr 12, 2018

Answer:

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

Explanation:

First, get the equation into the form #y=mx+c#
#3y=5x-3#
#y=5/3x-1#

The gradient of the perpendicular line is the negative reciprocal of the original line. The gradient of the original line is #5/3#, so the gradient of the perpendicular line is #-3/5#

Put this into the equation #y=mx+c#

#y=-3/5x+c#

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

#5=-3/5(-1)+c#
#5=3/5+c#
#c=22/5#

The equation of the line is #y=-3/5x+22/5#

Now for graphing.
You know the line passes through the point #(-1,5)#. Plot this point.
You know that the y-intercept is #(0,22/5)#. Plot this point.
The gradient of the line is #-3/5#, 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.