SOLUTION
The given points are: (-7,-3) and (-1,-1)
The slope of the line segment is:
[tex]\begin{gathered} m=\frac{-1-(-3)}{-1-(-7)} \\ m=\frac{-1+3}{-1+7} \\ m=\frac{2}{6} \\ m=\frac{1}{3} \end{gathered}[/tex]Recall that the product if solpes of perpendicular line give -1.
The the slope of the perpendicula bisector is:
[tex]\begin{gathered} m_1=\frac{1}{\frac{1}{3}} \\ m=3 \end{gathered}[/tex]Therefore the slope of the perpendicular bisector is -3.
Recall that the perpendicular bisector passes through the center of a line segment.
Hence the perpendicular bisector will pass through:
[tex]\begin{gathered} (\frac{-7-1}{2},\frac{-3-1}{2}) \\ =(\frac{-8}{2},\frac{-4}{2}) \\ =(-4,-2) \end{gathered}[/tex]Using the point slope form, the equation of the perpendicular bisector is:
[tex]\begin{gathered} y-(-2)=-3(x-(-4)) \\ y+2=-3x-12 \\ y=-3x-12-2 \\ y=-3x-14 \end{gathered}[/tex]Therefore the required equation is:
[tex]y=-3x-14[/tex]