Determining the image of a point reflected on an arbitrary line

Question

I need to determine the reflections of several points across different lines to solve a problem in electrostatics. So far, my approach has been to determine the equations of the line-segments uniting the points (A, A'), the equation for the axis of reflection, the point of intersection of the two (X) and adding to my initial point (A) the vector: $2\vec{AX}$.

However, since I need to repeat this 5 times, I was wondering - is there a more "systematic" way to determine the images of points under reflection?

Answer 1

You can instead use a reflection matrix: $$R(\theta)=\begin{pmatrix} \cos(2\theta) & \sin(2\theta)\\ \sin(2\theta) & -\cos(2\theta) \end{pmatrix}$$

where $\theta$ is the angle that axis of reflection makes with the x-axis (assuming it crosses the origin, if not you must first add the coordinates of the point at which it crosses the y-axis - which corresponds to making a translation, followed by a reflection.

So the coordinates of the reflected point will be $\vec r_0' = \vec r_0 \cdot R(\theta)$, where $\vec r_0$ are the coordinates of your first point.

Geogebra has a neat online app that allows you to calculate the reflection matrix and check your results.

Answer 2

You’re already doing something systematic, but you can certainly simplify it.

If the lines go through the origin, then there’s a standard formula that you can use to compute the reflection of the point $\mathbf p$ that’s based on the idea that you obtain the reflection by reversing the component of $\mathbf p$ that’s perpendicular to the reflection axis. Let $\mathbf n$ be a normal to the reflection axis. Then, after a bit of algebraic manipulation, you can find that the reflection of $\mathbf p$ is $$\mathbf p - 2{\mathbf n\cdot\mathbf p\over\mathbf n\cdot\mathbf n}\mathbf n.\tag1$$ You can extract $\mathbf n$ from the Cartesian equation of the reflection axis. If the reflection axis is instead defined by its direction $\mathbf d$, the same idea can be used to obtain a similar formula that uses this direction vector instead: $$2{\mathbf d\cdot\mathbf p\over\mathbf d\cdot\mathbf d}\mathbf d-\mathbf p.\tag2$$

If the reflection axis doesn’t pass through the origin, then you can always translate by a known point on the axis, use either of the above reflection formulas, and translate back. Alternatively, we can come up with a similar-looking formula using the Cartesian equation $ax+by+c=0$ of the reflection axis directly: Setting $\mathbf n=(a,b)$, we have $$\mathbf p-2{\mathbf n\cdot\mathbf p+c\over \mathbf n\cdot\mathbf n}\mathbf n.\tag3$$ If you need to reflect many points over the same line, it’s possible to rewrite all of these expressions as the product of a fixed matrix and $\mathbf p$ (using homogeneous coordinates for the last one), and then perform the reflections “in bulk.”