# How do you write an equation of a line passing through (5, -3), perpendicular to  y=6x + 9?

Jul 6, 2017

See a solution process below:

#### 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.

$y = \textcolor{red}{6} x + \textcolor{b l u e}{9}$

The slope of this line is: $\textcolor{red}{m = 6}$

Let's call the slope of a perpendicular line: ${m}_{p}$

The formula for the slope of a perpendicular line is: ${m}_{p} = - \frac{1}{m}$

Substituting gives us:

${m}_{p} = - \frac{1}{6}$

We can now use the point-slope formula to find an equation of the line from the problem. 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 $\left(\textcolor{red}{{x}_{1} , {y}_{1}}\right)$ is a point the line passes through.

Substituting the slope we calculated and the values from the point in the problem gives:

$\left(y - \textcolor{red}{- 3}\right) = \textcolor{b l u e}{- \frac{1}{6}} \left(x - \textcolor{red}{5}\right)$

$\left(y + \textcolor{red}{3}\right) = \textcolor{b l u e}{- \frac{1}{6}} \left(x - \textcolor{red}{5}\right)$

If we want the equation in slope-intercept form we can solve for $y$:

$y + \textcolor{red}{3} = \left(\textcolor{b l u e}{- \frac{1}{6}} \times x\right) - \left(\textcolor{b l u e}{- \frac{1}{6}} \times \textcolor{red}{5}\right)$

$y + \textcolor{red}{3} = - \frac{1}{6} x - \left(- \frac{5}{6}\right)$

$y + \textcolor{red}{3} = - \frac{1}{6} x + \frac{5}{6}$

$y + \textcolor{red}{3} - 3 = - \frac{1}{6} x + \frac{5}{6} - 3$

$y + 0 = - \frac{1}{6} x + \frac{5}{6} - \left(\frac{6}{6} \times 3\right)$

$y = - \frac{1}{6} x + \frac{5}{6} - \frac{18}{6}$

$y = \textcolor{red}{- \frac{1}{6}} x - \textcolor{b l u e}{\frac{13}{6}}$