To prove that $\mathrm{OX}=\mathrm{OX}\text{'}=p$, I drew a line XX', which intersects and is parallel to the line $y=x\tan \theta$ (call this point of intersection D). But $\mathrm{DX}\text{'}=\mathrm{DX}$ and so $\mathrm{ODX}$ and $\mathrm{ODX}\text{'}$ are two right-angled triangles of the same size, so $\mathrm{OX}=\mathrm{OX}\text{'}=p$. By the same argument I drew lines $\mathrm{OP}$ and $\mathrm{OP}\text{'}$, and so got right-angled triangles again, so $\mathrm{OP}=\mathrm{OP}\text{'}=q$.

By looking at the right-angled triangle $\mathrm{OAX}\text{'}$, with the angle at $O$ being $2\theta$, I knew that:

${\displaystyle \cos 2\theta =\frac{\mathrm{OA}\text{'}}{\mathrm{OX}}=\frac{\mathrm{OA}\text{'}}{p}}$

and so $\mathrm{OA}\text{'}=p\cos 2\theta$.

I then looked at the right-angled triangle $X\text{'}\mathrm{BP}$, and since the angle at $X$ is $2\theta$, $\mathrm{BP}\text{'}=q\sin 2\theta$. By applying Pythagoras' Theorem to the right-angled triangle $\mathrm{OAX}$, $\mathrm{AX}\text{'}=p\sin 2\theta$. Finally, applying Pythagoras' Thereom to the triangle $X\text{'}\mathrm{BP}\text{'}$ I found $\mathrm{BX}\text{'}=q\cos 2\theta$.

Looking at the change in X co-ordinates, I found $P\text{'}=(p\cos 2\theta +q\sin 2\theta ,p\sin 2\theta -q\cos 2\theta )$.

So the matrix for the reflection would be:

${\displaystyle T=\left(\begin{array}{cc}\hfill \cos 2\theta \hfill & \hfill \sin 2\theta \hfill \\ \hfill \sin 2\theta \hfill & \hfill -\cos 2\theta \hfill \end{array}\right)}$