College Algebra, Chapter 2, 2.4, Section 2.4, Problem 38

Find an equation of the line that pass through $\displaystyle (-2, -11)$ and perpendicular to the line passing through $(1,1)$ and $(5, -1)$.

First, we get determine the perpendicular line by using the two point form.


$\begin{equation} \begin{aligned} y - y_1 =& \left( \frac{y_2 - y_1}{x_2 - x_1} \right) (x - x_1) \\ \\ y - 1 =& \left( \frac{-1 - 1}{5 - 1} \right) (x - 1) \\ \\ y - 1 =& \left( \frac{-2}{4} \right) (x - 1) \\ \\ y - 1 =& \frac{-1}{2} (x - 1) \\ \\ y - 1 =& \frac{-1}{2} x + \frac{1}{2} \\ \\ y =& \frac{-1}{2} x + \frac{1}{2} \end{aligned} \end{equation}$


Therefore, the slope of the unknown line is $m = 2$.

By using Point Slope Form


$\begin{equation} \begin{aligned} y =& mx + b && \\ \\ y =& -2x + b && \text{Substitute } m =2 \\ \\ -11 =& 2(-2) + b && \text{Solve for } b \\ \\ b =& -7 && \end{aligned} \end{equation}$


Thus, the equation of the line is..

$y = 2x - 7$