Find the matrix which represents the reflection that maps triangle T2 onto triangle T3 http://filestore.aqa.org.uk/subjects/AQA-MFP1-QP-JUN13.PDF
Question 8(b)(i)
T2 Vertices $(1, 3)$, $(1, 6)$ and $(3, 3)$
Line: $\,\, y=\sqrt 3\, x$
How do you work this out? I am confused as to why this question only gives 2 marks? I have absolutely no idea how you would work this out?
In fact if anyone could go over the entire Question and how you work out each part I would really appreciate it, or at least guide me to a place I can learn how I've looked everywhere and cannot find anything. Regards Tom
 A: It's  an instance of the general problem: find the matrix of a reflection with respect to a line through the origin, that has a polar angle $\theta$. In this problem, $\theta=\frac\pi3$.
Change orthonormal basis: take as unit vectors the unit vector on the line, and the unit vector that makes an angle of $\frac\pi2$ with the first one. The change of basis matrix is then $$P=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$
(its column vectors are the coordinates of the new unit vectors). It is the matrix for the rotation of angle $\theta$ around the origin.
Now let $X$ be the column vector of coordinates of a point in the canonical basis, $X'$ its vector of coordinates in the new basis. they're linked through the relation $X=PX'$ The matrices $A$ and $A'$ of any linear transformation in the old and new bases respectively are linked through:
$$A'=P^{-1}AP\iff A=PA'P^{-1}.$$
The reason for this formula is that, if a point $X$ has reflection $Y$ as  (in the canonical base), then  $Y=AX$. Now, this can be rewritten in function of the coordinates $X'$ and $Y'$ in the new basis: $PY'=APX'$, hence $Y'=P^{-1}APX'=(P^{-1}AP)X'$, which proves the formula.
Now in the new basis the reflection is but the reflection with respect to the first axis, so that 
$$A'=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$
Also, $P^{-1}$ is the matrix of the rotation of angle $-\theta$ around the origin:
$$P^{-1}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$
Using  the duplication formulae in trigonometry, we get after some computations:
$$A=\begin{bmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{bmatrix}.\qquad \mathrm{Here:} \quad A=\begin{bmatrix}-\dfrac12&\dfrac{\sqrt3}2\\\dfrac{\sqrt3}2&\dfrac12\end{bmatrix}.$$
Some computation details using the slope $\boldsymbol{m=\sqrt 3}$:
Since $m=\tan\theta\,$ and $-\frac\pi2<\theta <\frac\pi2$, we have:
$\begin{cases}\cos\theta =\frac1{\sqrt{1+m^2}}\\[0.5ex]\sin\theta =\frac m{\sqrt{1+m^2}}\end{cases}$, so that: 
\begin{gather*} 
P=\frac1{\sqrt{1+m^2}}\begin{bmatrix}1&-m\\m&1\end{bmatrix},\quad P^{-1}=\frac1{\sqrt{1+m^2}}\begin{bmatrix}1&m\\-m&1\end{bmatrix}\\[1ex]
A=\dfrac{1}{1+m^2}\begin{bmatrix}1-m^2&2m\\
2m&m^2-1\end{bmatrix}.
\end{gather*}
