# What is the equation of the line that passes through the point (4,-5) and is perpendicular to 2x-5y= -10?

Oct 20, 2015

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

#### Explanation:

Rewrite the equation of the line we must be perpendicular to as $y = \frac{2 x + 10}{5} = \frac{2}{5} x + 2$. This is the slope-intercept form, and indeed we can see that the slope is $m = \frac{2}{5}$, and the intercept is $q = 2$ (even if we don't care about it in this specific case).

A line with slope $n$ is perpendicular to a line with slope $m$ if and only if the following equation holds:

$n = - \frac{1}{m}$.

In our case, the slope must be $- \frac{1}{\frac{2}{5}} = - \frac{5}{2}$.

So, now we know everything we need, since the slope and a known point identify a line uniquely: we can find the equation with the formula

$y - {y}_{0} = m \left(x - {x}_{0}\right)$, if $m$ is the slope of the line and $\left({x}_{0} , {y}_{0}\right)$ is the known point. Plugging the values, we have

$y + 5 = - \frac{5}{2} \left(x - 4\right)$, which we can adjust into

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