# How do you find the equation of a line that is perpendicular to y=2/3x-4 and goes through point (6,-2)?

Jan 29, 2015

The answer is: $y = - \frac{3}{2} x + 7$

The equation of a line, written in explicit form is:

$y = m x + q$

where

$m$ is the slope of the line and
$q$ is the y-intercept.

The formula that gives the equation of a line $r$ if its slope, ${m}_{r}$, and a point $P$ that $\in r$ are known is:

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

Two lines $r \mathmr{and} s$ are perpendicular if ${m}_{r} = - \frac{1}{m} _ s$.

So the slope of the requested line is: ${m}_{r} = - \frac{1}{\frac{2}{3}} = - \frac{3}{2}$.

The equation of the line is:

$y - \left(- 2\right) = - \frac{3}{2} \left(x - 6\right) \Rightarrow y + 2 = - \frac{3}{2} x + 9 \Rightarrow y = - \frac{3}{2} x + 7$.