College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 52

Find an equation of the line that passes through the point $(1,7)$ and is perpendicular to the line $x - 3y + 16 = 0$ in...
a.) Slope intercept form.
b.) General form.

If the line is perpendicular to $x - 3y + 16 = 0$, then their slopes must be the negative reciprocal of the other..

$\begin{equation} \begin{aligned} x - 3y + 16 &= 0 && \text{Add } 3y\\ \\ 3y &= x + 16 && \text{Divide both sides by } 3\\ \\ y &= \frac{x}{3} + \frac{16}{3} \end{aligned} \end{equation}$


By observation, the slope is $\displaystyle \frac{1}{3}$, so the slope of the perpendicular line is $m = -3$

Thus,

$\begin{equation} \begin{aligned} y &= mx + b \\ \\ y &= -3x + b \end{aligned} \end{equation}$


Solving for $b$ at point $(1,7)$

$\begin{equation} \begin{aligned} 7 &= -3(1) + b\\ \\ b &= 10 \end{aligned} \end{equation}$


Therefore, the equation of the line is... $y = -3x + 10$

b.) In general form

$\begin{equation} \begin{aligned} Ax + By + C &= 0 \\ \\ y &= -3x + 10 && \text{Add } 3x \text{ and subtract } 10\\ \\ 3x + y -10 &= 0 \end{aligned} \end{equation}$