The image of the pt. (x1,y1) in the mirror ax+by+c=0 is (x2,y2),P.T (x2-x1)/a=(y2-y1)/b=-2(ax1+by1+c)/a^2+b^2?

The mirror is along the line -x/(c/a) - y/(c/b) = 1 where x, y intercepts are (-(c/a) and -(c/b)

The midpoint say 'C' on the mirror for Normal Ray = (-c/2a, -c/2b)

Any ray from point say 'P' (x1,y1) to the mid point 'C' of the mirror will form image at 'P' (x2,y2)

Hence the slope of Ray CP and CQ will be equal

Hence slope CP = slope of CQ or ((y1+(c/2b))/((x1 +(c/2a)) = ((y2+(c/2b))/((x2 +(c/2a))

On simplification we get (x2-x1)/a = (y2 -y1)/2b

Also equating the perpendicular distance from P and Q to the line normal to the palne of the mirror we get (x2-x1)/a = -2(ax1 +by1+c)/(a^2 +b^2