Question

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

Answer 1

`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.