# How do you find equation of a line, L which passes through the point (3, -1) and perpendicular to the line with equation 9x - 3y = 2?

Dec 17, 2016

$y + = - \frac{1}{3} \left(x - 3\right)$ or $y = - \frac{1}{3} x$

#### Explanation:

To solve this we must first determine the slope of the given line by converting the equation to the slope-intercept form which is: $\textcolor{red}{y = m x - b}$ where $\textcolor{red}{m}$ is the slope.

$9 x - 9 x - 3 y = - 9 x + 2$

$0 - 3 y = - 9 x + 2$

$- 3 y = - 9 x + 2$

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

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

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

So the slope of the given line is $3$.

The slope of a perpendicular line is the negative inverse of the line give. So if $m$ is the slope of the given line the $- \frac{1}{m}$ is the slope of a perpendicular line.

So, in this case the slope of the perpendicular line is $- \frac{1}{3}$

We now have a slope and a point so we can use the point-slope to determine the equation for the perpendicular line.

The point-slope formula states: $\textcolor{red}{\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{red}{\left({x}_{1} , {y}_{1}\right)}$ is a point the line passes through.

Substituting the information we were given and calculated gives:

$y - - 1 = - \frac{1}{3} \left(x - 3\right)$

#y + 1 = -1/3x + 1/3 * 3

$y + 1 = - \frac{1}{3} x + 1$

or converting to the more familiar point-intercept form gives:

$y + 1 - 1 = - \frac{1}{3} x + 1 - 1$

$y + 0 = - \frac{1}{3} x + 0$

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