Answer
$y=\frac{3}{4}x-\frac{15}{4}$
Work Step by Step
We are given the points $A(-2,1)$ and $B(4,-7)$.
a) $\bf{Step\text{ }1}$
Find the midpoint of the line segment using $A(x_1,y_1)=(-2,1)$, $B(x_2,y_2)=(4,-7)$:
$$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)=\left(\dfrac{-2+4}{2},\dfrac{1+(-7)}{2}\right)=(1,-3)$$
b) $\bf{Step\text{ }2}$
Calculate the slope $m$ of $\overline{AB}$:
$$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-7-1}{4-(-2)}=-\dfrac{4}{3}.$$
$\bf{Step\text{ }3}$
Find the slope $m_1$ of the perpendicular bisector:
$$\begin{align*}
m_1&=-1\\
m_1&=-\dfrac{1}{m_1}=-\dfrac{1}{-\frac{4}{3}}=\dfrac{3}{4}.
\end{align*}$$
$\bf{Step\text{ }4}$
Use point-slope form:
$$\begin{align*}
y-(-3)&=m_1(x-1)\\
y+3&=\dfrac{3}{4}(x-1)\\
y&=\dfrac{3}{4}x-\dfrac{15}{4}.
\end{align*}$$
The equation of the perpendicular bisector is
$$y=\dfrac{3}{4}x-\dfrac{15}{4}.$$
