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

Aug 7, 2016

$\implies y = \frac{3}{2} x - \frac{19}{2}$

#### Explanation:

Let the given point be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(3 , - 5\right)$

Given equation;$\text{ } 2 x + 3 y = - 5$

Write this in standard form of: $y = m x + c$
where m is the gradient (slope).

$\textcolor{b r o w n}{2 x + 3 y = - 5} \textcolor{b l u e}{\text{ "->" } y = - \frac{2}{3} x - \frac{5}{3}}$

So for this line the gradient (m) is $- \frac{2}{3}$
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The gradient of the line perpendicular to this is $- \frac{1}{m}$

So $- \frac{1}{m} = + \frac{3}{2}$

Thus the perpendicular line equation is:$\text{ } y = \frac{3}{2} x + c$

We know that this line passes through the point ${P}_{2}$

So ${P}_{2} \to {y}_{2} = \frac{3}{2} {x}_{2} + c$

$\implies - 5 = \frac{3}{2} \left(3\right) + c$

$\implies c = - \frac{19}{2}$

$\implies y = \frac{3}{2} x - \frac{19}{2} \leftarrow \text{ perpendicular line}$