# How do you write the equation in point slope form given (-2,-2) perpendicular to y= -1/3x + 9?

Feb 14, 2017

$y = \textcolor{red}{3} x + \textcolor{b l u e}{4}$

#### Explanation:

The equation in the problem is in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y = \textcolor{red}{- \frac{1}{3}} x + \textcolor{b l u e}{9}$ so we know the slope of this line is $\textcolor{red}{m = - \frac{1}{3}}$

The slope of a perpendicular line (let's call it ${m}_{p}$) is the negative inverse of the slope of this line or ${m}_{p} = - \frac{1}{m}$

Therefore the slope of a perpendicular line is ${m}_{p} = 3$

We can now use the point-slope formula to find an equation for the line. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the point from the problem and the slope we determined gives:

$\left(y - \textcolor{red}{- 2}\right) = \textcolor{b l u e}{3} \left(x - \textcolor{red}{- 2}\right)$

$\left(y + \textcolor{red}{2}\right) = \textcolor{b l u e}{3} \left(x + \textcolor{red}{2}\right)$

We can now solve this for $y$ to put it into slope-intercept form:

$y + \textcolor{red}{2} = \left(\textcolor{b l u e}{3} \times x\right) + \left(\textcolor{b l u e}{3} \times \textcolor{red}{2}\right)$

$y + \textcolor{red}{2} = 3 x + 6$

$y + \textcolor{red}{2} - 2 = 3 x + 6 - 2$

$y + 0 = 3 x + 4$

$y = \textcolor{red}{3} x + \textcolor{b l u e}{4}$