# How do you find the equation, in standard form, of the line perpendicular to 2x + 3y = -5 and passing through (3, -5)?

Apr 5, 2015

The first information that we have is that we're looking for a line passing through $\left(3 , - 5\right)$. The family of lines passing through a given point $\left({x}_{0} , {y}_{0}\right)$ is $y - {y}_{0} = m \left(x - {x}_{0}\right)$, where $m$ is the slope of the line.

So, the lines passing through $\left(3 , - 5\right)$ are of the form

$y + 5 = {m}_{p} \left(x - 3\right)$, where i called the slope ${m}_{p}$ for perpendicular.

To find ${m}_{p}$, let's find the slope $m$ of the original line before: bringing into standard form the equation $2 x + 3 y = - 5$, we get $y = \setminus \frac{- 2 x - 5}{3} = - \frac{2}{3} x - \frac{5}{3}$. So, the original line has slope $- \frac{2}{3}$.

Now, two lines are perpendicular if their slopes $m$ and $m '$ are in the relation $m = - \frac{1}{m '}$.

From this relation, we have ${m}_{p} = \frac{3}{2}$, and the solution is thus

$y + 5 = \frac{3}{2} \left(x - 3\right)$

Here's a link where you can check that the two lines are perpendicular, and that the second one passes through $\left(3 , 5\right)$.