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..

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