lets say for example we want to reflect point (-6,-4) which is A

and lets say we want to reflect it across like line y=-4x-4

Well you could probably do it graphically.

or

You could say that the gradient of the line is -4 so the gradient of the perpendicular is 1/4

The equation of the perpendicular through (-6,-4) is

y=0.25x+k

-4=0.25*-6+k

-4=-1.5+k

k=-2.5

so

y=0.25x-2.5

simulaneous solution of the two lines

\(y=-4x-4 \\ y=0.25x-2.5\\ -4x-4=0.25x-2.5\\ -1.5=4.25x\\ x=-6/17\\ y=-2\frac{10}{17}\\ (\frac{-6}{17},-2\frac{10}{17})\)

Now you can find the distance between those two points, then find the other point on the perpendicular that is on other side of the reflection line and the same distance away as the original point was.

These numbers are horrible. Did you make this question up?