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

Apr 12, 2018

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

#### Explanation:

First, get the equation into the form $y = m x + c$
$3 y = 5 x - 3$
$y = \frac{5}{3} x - 1$

The gradient of the perpendicular line is the negative reciprocal of the original line. The gradient of the original line is $\frac{5}{3}$, so the gradient of the perpendicular line is $- \frac{3}{5}$

Put this into the equation $y = m x + c$

$y = - \frac{3}{5} x + c$

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

$5 = - \frac{3}{5} \left(- 1\right) + c$
$5 = \frac{3}{5} + c$
$c = \frac{22}{5}$

The equation of the line is $y = - \frac{3}{5} x + \frac{22}{5}$

Now for graphing.
You know the line passes through the point $\left(- 1 , 5\right)$. Plot this point.
You know that the y-intercept is $\left(0 , \frac{22}{5}\right)$. Plot this point.
The gradient of the line is $- \frac{3}{5}$, 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.