# How do you write the equation of line passes through (4, -5), and is perpendicular to 2x-5y= -10?

Sep 2, 2016

$5 x + 2 y = 10$

#### Explanation:

Let us write the equation of line 2x-5y=-10 in slope intercept form i.e.

$- 5 y = - 2 x - 10$ or $y = \frac{2}{5} x + 2$

i.e. its slope is $\frac{2}{5}$. Hence the slope of a line perpendicular to it is $\frac{- 1}{\frac{2}{5}} = - \frac{5}{2}$.

Now equation of a line passing through appoint $\left({x}_{1} , {y}_{1}\right)$ and having a slope $m$ is given by

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$ and

so the equationn of line passing through $\left(4 , - 5\right)$ and having a slope of $- \frac{5}{2}$ is

$\left(y - \left(- 5\right)\right) = - \frac{5}{2} \left(x - 4\right)$ or

y+5=-5/2x+(5×4)/2# or

$y + 5 = - \frac{5}{2} x + 10$ or

$2 y + 10 = - 5 x + 20$ or

$5 x + 2 y = 10$