Answer
$y=\frac{1}{3}x+\frac{20}{3}$
Work Step by Step
We are given the points $A(0,0)$ and $B(-4,12)$.
a) $\bf{Step\text{ }1}$
Find the midpoint of the line segment using $A(x_1,y_1)=(0,0)$, $B(x_2,y_2)=(-4,12)$:
$$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)=\left(\dfrac{0+(-4)}{2},\dfrac{0+12}{2}\right)=(-2,6)$$
b) $\bf{Step\text{ }2}$
Calculate the slope $m$ of $\overline{AB}$:
$$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{12-0}{-4-0}=-3.$$
$\bf{Step\text{ }3}$
Find the slope $m_1$ of the perpendicular bisector:
$$\begin{align*}
m_1&=-1\\
m_1&=-\dfrac{1}{m_1}=-\dfrac{1}{-3}=\dfrac{1}{3}.
\end{align*}$$
$\bf{Step\text{ }4}$
Use point-slope form:
$$\begin{align*}
y-6&=m_1(x-(-2))\\
y-6&=\dfrac{1}{3}(x+2)\\
y&=\dfrac{1}{3}x+\dfrac{20}{3}.
\end{align*}$$
The equation of the perpendicular bisector is
$$y=\frac{1}{3}x+\frac{20}{3}.$$
