Answer
$2x+y=-27$
Work Step by Step
Using $\left( \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the midpoint of the line segment joining the points $
(-2,6) \text{ and } (-22,-4)
$ is
\begin{array}{l}\require{cancel}
\left( \dfrac{-6+(-22)}{2},\dfrac{6+(-4)}{2} \right)
\\\\
\left( \dfrac{-6-22}{2},\dfrac{6-4}{2} \right)
\\\\
\left( \dfrac{-28}{2},\dfrac{2}{2} \right)
\\\\
\left( -14,1 \right)
.\end{array}
Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, then \begin{array}{l}\require{cancel}
m=\dfrac{6-(-4)}{-2-(-22)}
\\\\
m=\dfrac{6+4}{-2+22}
\\\\
m=\dfrac{10}{20}
\\\\
m=\dfrac{1}{2}
.\end{array}
Taking the negative reciprocal of $m$, then the slope of the perpendicular bisector is $m_p=
-2
.$
Using $
(-14,1)
$ and $m_p=
-2
,$ the equation of the perpendicular bisector is
\begin{array}{l}\require{cancel}
y-1=-2(x-(-14))
\\\\
y-1=-2(x+14)
\\\\
y-1=-2x-28
\\\\
2x+y=-28+1
\\\\
2x+y=-27
.\end{array}
