Intermediate Algebra, Chapter 3, Summary Exercises, Section Summary Exercises, Problem 14

Write an equation of the line "through $(4,-2)$ and perpendicular to the line through $(3,7)$ and $(5,6)$".

(a) In slope-intercept form

Using the Slope Formula

$\displaystyle m = \frac{6-7}{5-3} = - \frac{1}{2}$

The slope is $\displaystyle - \frac{1}{2}$. We know that if the lines are perpendicular, the product of their slope is $-1$. So the slope of the perpendicular line is $2$.

Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
y - (-2) =& 2(x - 4)
&& \text{Substitute } x = 4, y = -2 \text{ and } m = 2
\\
y + 2 =& 2x - 8
&& \text{Distributive Property}
\\
y =& 2x - 8 - 2
&& \text{Subtract each side by $2$}
\\
y =& 2x - 10
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$



(b) In standard form


$
\begin{equation}
\begin{aligned}

y =& 2x - 10
&& \text{Slope Intercept Form}
\\
-2x + y =& -10
&& \text{Standard Form}
\\
\text{or} &
&&
\\
2x - y =& 10
&&

\end{aligned}
\end{equation}
$