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# Rots and Refs

Bob sent us his solution:

Firstly I multiplied the matricies to find the new point. Using the trigonometric identities to simplify, I got the new point as $(r\cos(\theta+\phi),r\sin(\theta+\phi))$. This meant that we had rotated the point $\phi$ degrees anticlockwise.

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

By looking at the right-angled triangle $OAX'$, with the angle at $O$ being $2\theta$, I knew that: $$\cos2\theta=\frac{OA'}{OX}=\frac{OA'}{p}$$ and so $OA'=p\cos2\theta$.

I then looked at the right-angled triangle $X'BP$, and since the angle at $X$ is $2\theta$, $BP'=q\sin2\theta$. By applying Pythagoras' Theorem to the right-angled triangle $OAX$,$AX'=p\sin2\theta$. Finally, applying Pythagoras' Thereom to the triangle $X'BP'$ I found $BX'=q\cos2\theta$.

Looking at the change in X co-ordinates, I found $P'=(p\cos2\theta+q\sin2\theta, \; p\sin2\theta-q\cos2\theta)$.

So the matrix for the reflection would be:

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Bob sent us his solution:

Firstly I multiplied the matricies to find the new point. Using the trigonometric identities to simplify, I got the new point as $(r\cos(\theta+\phi),r\sin(\theta+\phi))$. This meant that we had rotated the point $\phi$ degrees anticlockwise.

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

By looking at the right-angled triangle $OAX'$, with the angle at $O$ being $2\theta$, I knew that: $$\cos2\theta=\frac{OA'}{OX}=\frac{OA'}{p}$$ and so $OA'=p\cos2\theta$.

I then looked at the right-angled triangle $X'BP$, and since the angle at $X$ is $2\theta$, $BP'=q\sin2\theta$. By applying Pythagoras' Theorem to the right-angled triangle $OAX$,$AX'=p\sin2\theta$. Finally, applying Pythagoras' Thereom to the triangle $X'BP'$ I found $BX'=q\cos2\theta$.

Looking at the change in X co-ordinates, I found $P'=(p\cos2\theta+q\sin2\theta, \; p\sin2\theta-q\cos2\theta)$.

So the matrix for the reflection would be:

\mathbf{T}= \left(
\begin{array}{cc} cos 2\theta & sin2\theta\\ sin2\theta &
-\cos2\theta \end{array} \right)

You may also like to look at the
problem ' The Matrix ' from July 2003
and its solution for an explanation of how a transformation of the
plane is given by a matrix and how you can find the image of a
point by multiplying its vector by the matrix of the
transformation.