Find a formula in terms of $a, b$ and $c$, for reflection in an arbitrary line $ax + by + c = 0 \in \mathbb{R}^{2}$.

Question

I am very close to finding the formula. I have the product of the slopes which is $bx-ay + aq - bp = 0$ and the equations $|ax+by+c| = |ap+bq+c|$.

I need to find the values for $p$ and $q$.

Any suggestions?

Answer 1

When taking the approach that you’ve chosen, it’s better to work with signed distances: if the distance of $P_1=(x_1,y_1)$ from the reflection line $L$ is $d$, then the distance of its reflection will be $-d$. You can then construct a line parallel to the reflector at this distance from it and compute the intersection of this line with the perpendicular through $P_1$.

In more detail, the signed distance of $P_1$ from the line is $d={ax_1+by_1+c \over \sqrt{a^2+b^2}}$. The sign of this value tells you whether this distance is measured in the direction of the normal $\langle a,b \rangle$ or in the opposite direction. If you add $ax_1+by_1+c$ to $L$’s equation, you get an equation of a line that’s at a signed distance of $-d$ from $L$, i.e., just as far from $L$ as $P_1$, but on the opposite side. (You can verify this by computing the distance of $P_1$ from this line: it will be $2d$.) You can find an equation of the perpendicular through $P_1$ by using the point-normal form: $(b,-a)\cdot(x,y)=(b,-a)\cdot(x_1,y_1)$. Find the intersection of these lines to find the image of $P_1$, and you will have a formula for this reflection.

There’s are somewhat simpler ways to go about this, though. For instance, if you translate and rotate so that $L$ becomes the $x$-axis, then the reflection is simply $x\mapsto -x$. Make this substution and then rotate and translate back.