# What is the equation of the line perpendicular to y =3x- 7 that contains (6, 8)?

Dec 29, 2016

$\left(y - 8\right) = - \frac{1}{3} \left(x - 6\right)$

or

$y = - \frac{1}{3} x + 10$

#### Explanation:

Because the line given in the problem is in the slope intercept form we know the slope of this line is $\textcolor{red}{3}$

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 color(blue)(b is the y-intercept value.
This is a weighted average problem.

Two perpendicular lines have a negative inverse slope of each other.

The line perpendicular to a line with slope $\textcolor{red}{m}$ has a slope of $\textcolor{red}{- \frac{1}{m}}$.

Therefore, the line we are looking for has a slope of $\textcolor{red}{- \frac{1}{3}}$.

We can now use the point-slope formula to find the equation of the line we are looking for.

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.

We can substitute the slope we calculate and the point we were given to give the equation we are looking for:

$\left(y - \textcolor{red}{8}\right) = \textcolor{b l u e}{- \frac{1}{3}} \left(x - \textcolor{red}{6}\right)$

If we want to put this in slope-intercept form we can solve for $y$:

y - color(red)(8) = color(blue)(-1/3)x - (color(blue)(-1/3) xx color(red)(6)))

$y - \textcolor{red}{8} = \textcolor{b l u e}{- \frac{1}{3}} x - \left(- 2\right)$

$y - \textcolor{red}{8} = \textcolor{b l u e}{- \frac{1}{3}} x + 2$

$y - \textcolor{red}{8} + 8 = \textcolor{b l u e}{- \frac{1}{3}} x + 2 + 8$

$y - 0 = \textcolor{b l u e}{- \frac{1}{3}} x + 10$

$y = - \frac{1}{3} x + 10$