In preparing for a problem-solving session, I came across this lovely problem from the MAT:

The reflection of point $(1,0)$ in the line $y=mx$ has the coordinates:

A: $\left( \frac{m^2+1}{m^2-1}, \frac{m}{m^2-1}\right)$ B: $(1,m)$ C: $(1-m,m)$ D: $\left( \frac{1-m^2}{1+m^2}, \frac{2m}{1+m^2}\right)$ E: $(1-m^2, m)$

You might want to have a little think about how you’d approach the question, and I’ll run through three options below the line. Spoilers ahead!

Elimination method

Given a multiple choice question, I quite often try to remove impossible answers before I do anything else. Considering simple cases is a valid strategy!

So, what if $m=0$? The point remains in place. That eliminates option A, which maps to $(-1,0)$.

Another simple case: what if $m$ is very big and positive? Then the reflection will lie near to $(-1,0)$, with a small positive $y$ coordinate. In one stroke, that rules out B, C and E, leaving only D as a possibility!

That’s a perfectly acceptable way to tackle a question in a multiple choice exam. But it’s not the nicest way to tackle it. Let’s look at two alternatives.

Matrices

If you know your matrix transformations, this is fairly straightforward: if a line of reflection has the equation $y = x\tan(\theta)$, then the matrix corresponding to it is $\mattwotwo {\cos(2\theta)}{\sin(2\theta)}{-\sin(2\theta)}{\cos(2\theta)}$.

Applying that to $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ gives $\colvectwo {\cos(2\theta)}{-\sin(2\theta)}$.

(Again, D is the only option that can possibly match that for all $\theta$, but suppose we don’t know the answer already.)

Now, we have $m = \tan(\theta)$, and I’m going to go via $\tan(2\theta) = \frac{2m}{1-m^2}$, although other options are available.

Thinking about a triangle with legs of $2m$ and $1-m^2$, the hypotenuse has to be $1 + m^2$, which I’ll leave as a short exercise – and the form of option D drops straight out.

Circle theorems

“Greetings, sensei.”

“The point is on the unit circle centred on the origin. The line of reflection contains a diameter of the circle. The chord through the point perpendicular to that diameter has the crossing point as its midpoint, so the image of the point also lies on the circle.”

“This would be easier with a diaplease don’t impale me on that very sharp pencil.”

“The only one of the options that can possibly form a Pythagorean triple is D. It turns out you can use that to generate all such triples.”

“Thank you, sensei.”

Did you have another approach? Please let me know in the comments!