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

Aug 10, 2016

#### Answer:

$y = - \frac{3}{2} x - 6 \frac{1}{2}$

#### Explanation:

We need the slope of the line we have been given before we can find the slope of the new line.

Change $2 x - 3 y = 3$ into the form $y = m x + c$

$2 x - 3 = 3 y \text{ "rArr" } 3 y = 2 x - 3$
$y = \frac{2}{3} x - 1. \text{ } {m}_{1} = \frac{2}{3}$.

The slope perpendicular to this is ${m}_{2} = - \frac{3}{2}$
We have the slope , $\left(- \frac{3}{2}\right)$and a point, $\left(- 3 , - 2\right)$
substitute the values into

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

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

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

$y = - \frac{3}{2} x - \frac{9}{2} - 2$

$y = - \frac{3}{2} x - 6 \frac{1}{2}$