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..

$y = 2x - 7$