check the picture below.
so, those are the points, now, the line of symmetry, the red one, is the line that will cut the vertex is half and will be perpendicular to the opposite side.
now, the slope of the black line we can just get it off the grid, notice the rise and run, rise/run = 6/6 = 1.
now a perpendicular line to that one will have a negative reciprocal slope, thus
[tex] \bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{1\implies \cfrac{1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{1}\implies -1}} [/tex]
so the slope of that perpendicular line will be -1, so we're really looking for the equation of a line whose slope is -1 and runs through -6,3.
[tex] \bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad \qquad \qquad
slope = m\implies -1
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=-1[x-(-6)]
\\\\\\
y-3=-1(x+6)\implies y-3=-x-6\implies y=-x-3 [/tex]
