# How do you find the equation of the line in slope-intercept form that is perpendicular to the line -4x - 3y = 9 and passes through (12, -7)?

Aug 26, 2015

$y = \frac{3}{4} x - 16$

#### Explanation:

$- 4 x - 3 y = 9$
can be re-written in slope intercept form as:
$y = - \frac{4}{3} x - 3$
and therefore has a slope of $\left(- \frac{4}{3}\right)$

Lines which are perpendicular to each other have slopes that are the negative reciprocal of one another.

Therefore any line perpendicular to $- 4 x - 3 y = 9$ has a slope of
$m = \frac{3}{4}$

Since the desired line has a slope of $\frac{3}{4}$ and passes through $\left(12 , - 7\right)$ we can write it's equation in slope-point form as:
$\left(y - \left(- 7\right)\right) = \frac{3}{4} \left(x - 12\right)$

Simplifying:
$y + 7 = \frac{3}{4} x - 9$
or
$y = \frac{3}{4} x - 16 \textcolor{w h i t e}{\text{XXXX}}$which is slope-intercept form.