Problem Statement:-
A ray of light is sent along the line $\ell_1\equiv x-2y+5=0$. Upon reaching the line $\ell_2\equiv3x+2y+7=0$ the ray is reflected from it. Find the equation of the line containing the reflected ray.
I had initially thought of attempting this problem by considering the signed angle of the incident line w.r.t the normal of the mirror line and then finding a line which would have the same signed angle in magnitude but opposite in direction w.r.t the normal and passing through the intersection of the point of the mirror and the incident line.
But this seemed too long for me and there were too many chances of some calculation errors.
So I came up with this attempt of mine.
Attempt at a solution:-
As we know that the reflection of a point w.r.t a line is given by the following formula:-
$$\frac{x_m-x}{a}=\frac{y_m-y}{b}=-2\left(\frac{ax+by+c}{a^2+b^2}\right)$$
where $(x_m,y_m)$ is the reflection/image of the point $(x,y)$ about the line $ax+by+c=0$
So, if we consider any point $(x,y)$ on the line $\ell_1$, then its reflection/image about the line $\ell_2$ will be given as:-
$$x_m=-2\times 3\left(\frac{3x+2y+7}{13}\right)+x,\;\; y_m=-2\times 2\left(\frac{3x+2y+7}{13}\right)+y$$
On substituting $(x_m,y_m)$ in the line $\ell_1=0$, we get the correct reflection line which is $19x-22y+79=0$
So why is it that on substituting the the image coordinates to the equation of the incident line we get the equation of the reflected line.