# Find the equation, in standard form, of the line perpendicular to 2x+3y=−5 and passing through (3, -5)

Find the equation, in standard form, of the line perpendicular to 2x+3y=−5 and passing through (3, -5)
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Marie Horn
The first information that we have is that we're looking for a line passing through (3,−5). 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 (3,−5) 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 2x+3y=−5, we get $y=\frac{-2x-5}{3}=-\frac{2}{3}x-\frac{5}{3}$. So, the original line has slope $-\frac{2}{3}$.
Now, two lines are perpendicular if their slopes m and m' are in the relation $m=-\frac{1}{m\prime }$.
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 (3,5).