The mirror is the line 3x - 2y = 2. Find the image of the point P(-3, 1) in it, and the distance…

The bisectors of two lines are $\dfrac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm\dfrac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$. With $c_1, c_2 > 0$, the plus sign goes through the angle that holds the origin, which is acute when $a_1a_2 + b_1b_2 < 0$. The image of $(x_1, y_1)$ in $ax + by + c = 0$ is $(x_1 - 2at,\ y_1 - 2bt)$ with $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$. A ray from $A$ to $B$ by way of a mirror strikes where the line from $A$ to the image of $B$ meets the mirror.