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}
$