# How do you write an equation of a line passing through (3, -5), perpendicular to 5x+2y=10?

Jul 4, 2016

It is $y = \frac{2}{5} x - \frac{31}{5}$.

#### Explanation:

First of all we put the line $5 x + 2 y = 10$ in the form $y = m x + q$.

$5 x + 2 y = 10$

$2 y = - 5 x + 10$

$y = - \frac{5}{2} x + 5$.

Now it is easy to find all the lines perpendicular to this. It is enough to take the opposite inverse of the slope $m \to - \frac{1}{m}$. In this case $m = - \frac{5}{2}$ then the slope of the perpendicular lines is $\frac{2}{5}$ and the equation is

$y = \frac{2}{5} x + q$.

To find $q$ we impose the passage for the point $\left(3 , - 5\right)$

$- 5 = \frac{2}{5} \cdot 3 + q$

$- 5 = \frac{6}{5} + q$

$- 5 - \frac{6}{5} = q$

$- \frac{31}{5} = q$.

The equation of the perpendicular is finally

$y = \frac{2}{5} x - \frac{31}{5}$.