College Algebra, Chapter 2, 2.4, Section 2.4, Problem 36

Find an equation of the line that pass through $\displaystyle \left( \frac{1}{2}, \frac{-2}{3} \right)$ and perpendicular to the line $4x - 8y = 1$.

If the line is perpendicular to $4x - 8y = 1$, then its slope is equal to the negative reciprocal of the other.


$
\begin{equation}
\begin{aligned}

4x - 8y =& 1
&&
\\
\\
8y =& 4x - 1
&& \text{Add $8y$ and subtract } 1
\\
\\
y =& \frac{4}{8} x - \frac{1}{8}
&& \text{Divide by } 8
\\
\\
y =& \frac{1}{2}x - \frac{1}{8}
&&

\end{aligned}
\end{equation}
$


By observation, the slope of the line perpendicular line to $4x - 8y = 1$ is $m = -2$.

By using Point Slope Form


$
\begin{equation}
\begin{aligned}

y =& mx + b
&&
\\
\\
y =& -2x + b
&& \text{Substitute } m =-2
\\
\\
\frac{-2}{3} =& -2 \left( \frac{1}{2} \right) + b
&& \text{Solve for } b
\\
\\
\frac{-2}{3} =& -1 + b
&& \text{Simplify}
\\
\\
b =& \frac{1}{3}
&&

\end{aligned}
\end{equation}
$


Thus, the equation of the line is..

$y = -2x + \frac{1}{3}$