# How do you write an equation of a line given (-5, -8) and is perpendicular to 10x – 6y = -11?

Jul 22, 2017

See a solution process below:

#### Explanation:

The equation in the problem is in Standard Linear form. The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$

Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: $m = - \frac{\textcolor{red}{A}}{\textcolor{b l u e}{B}}$

Given: $\textcolor{red}{10} x - \textcolor{b l u e}{6} y = \textcolor{g r e e n}{- 11}$

Then the slope is: $m = \frac{\textcolor{red}{- 10}}{\textcolor{b l u e}{- 6}} = \frac{5}{3}$

The formula for the slope of a perpendicular line, ${m}_{p}$ is:

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

Substituting gives: ${m}_{p} = \frac{- 1}{\frac{5}{3}} = - \frac{3}{5}$

Now we can use the point-slope formula to write and equation for the line to solve 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:

$\left(y - \textcolor{red}{- 8}\right) = \textcolor{b l u e}{- \frac{3}{5}} \left(x - \textcolor{red}{- 5}\right)$

$\left(y + \textcolor{red}{8}\right) = \textcolor{b l u e}{- \frac{3}{5}} \left(x + \textcolor{red}{5}\right)$

If necessary we can solve this equation for $y$ to put it 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.

$y + \textcolor{red}{8} = \left(\textcolor{b l u e}{- \frac{3}{5}} \cdot x\right) + \left(\textcolor{b l u e}{- \frac{3}{5}} \cdot \textcolor{red}{5}\right)$

$y + \textcolor{red}{8} = - \frac{3}{5} x - 3$

$y + \textcolor{red}{8} - 8 = - \frac{3}{5} x - 3 - 8$

$y + 0 = - \frac{3}{5} x - 11$

$y = \textcolor{red}{- \frac{3}{5}} x - \textcolor{b l u e}{11}$