## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| T is a reflection (in 2-D) and in any reflection any point on the mirror line remains invariant | B1 2.4 | Any point on the mirror line stays where it is |
| ...and so the mirror line must itself be a line of invariant points | B1 [2] 2.2a | ...so the mirror line is a line of invariant points. Accept "so the line of invariant points is the mirror line". If B0B0 then SC1 for any answer which is, in effect, a statement that the mirror line is an invariant line |

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## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For line of invariant points $A\mathbf{r} = \mathbf{r}$ | B1 1.1 | |
| $\frac{1}{13}\begin{pmatrix}5 & 12\\12 & -5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{13}\begin{pmatrix}5x+12y\\12x-5y\end{pmatrix}$ | M1 1.1 | Multiplying general point into **A**. Could be awarded for sight of $5x+12y$ or $12x-5y$ o.e. |
| $=\begin{pmatrix}x\\y\end{pmatrix} \Rightarrow \frac{1}{13}(5x+12y)=x$ or $\frac{1}{13}(12x-5y)=y$ | M1 1.1 | Equating and deriving an equation relating $x$ and $y$. Need to check that both equations give same straight line |
| $12y=8x$ (or $18y=12x$) $\Rightarrow y=\frac{2}{3}x$ or $y=\frac{2}{3}x+0$ | A1 [4] 1.1 | |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Line passes through $O \Rightarrow c=0$ | B1 | Used in the solution |
| $\frac{1}{13}\begin{pmatrix}5 & 12\\12 & -5\end{pmatrix}\begin{pmatrix}x\\mx\end{pmatrix}$ | M1 | Considering the matrix acting on a general point on the line $y=mx$ $(+c)$ |
| $=\frac{1}{13}\begin{pmatrix}5x+12mx\\12x-5mx\end{pmatrix}=\begin{pmatrix}x\\mx\end{pmatrix}$ | M1 | Multiplying and equating |
| $5x+12mx=13x$ and $12x-5mx=13mx \Rightarrow m=2/3$ so $y=\frac{2}{3}x$ or $y=\frac{2}{3}x+0$ | A1 | Need to check that both equations are satisfied by $m=2/3$ |

**Alternative method 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{13}\begin{pmatrix}5 & 12\\12 & -5\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix} = \frac{1}{13}\begin{pmatrix}5x+12mx+12c\\12x-5mx-5c\end{pmatrix}=\begin{pmatrix}x'\\y'\end{pmatrix}$ | M1 | Considering the matrix acting on a general point on the line $y=mx+c$ |
| $y'=mx'+c \Rightarrow \frac{1}{13}(12x-5mx-5c) = \frac{m}{13}(5x+12mx+12c)+c$ leading to $x(12m^2+10m-12)+c(12m+18)=0$ i.e. $2x(2m+3)(3m-2)+6c(2m+3)=0$ | M1 | Multiplying and substituting into $y=mx+c$ |
| Either $m=\frac{-3}{2}$, $c$ can be anything, or $m=\frac{2}{3}$ and $c=0$ | A1 | Finding two correct values of $m$ and no others (linking to $c$ not necessary here) |
| Single line $m=\frac{2}{3}$ and so $y=\frac{2}{3}x$ (perpendicular lines with $m=-\frac{3}{2}$ are invariant lines not invariant point lines) | A1 [4] | Convincing reason why $m=-3/2$ is rejected as a possibility |

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## Question 6(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{13}\begin{pmatrix}5 & 12\\12 & -5\end{pmatrix}\begin{pmatrix}1\\5\end{pmatrix}=\begin{pmatrix}5\\-1\end{pmatrix}$ [so $P'$ is $(5,-1)$] | M1 1.1 | SC: If answer given as a vector then maximum mark is M1 A1 A0. M1 could be awarded for solving $A\mathbf{x}=(5,1)$ to get $\mathbf{x}=(5,-1)$ |
| So required $x$-coord is $\frac{1}{2}(5+1)=3$... | A1 2.2a | |
| ...and required $y$-coord is $\frac{1}{2}(5+(-1))=2$ | A1 [3] 1.1 | |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of line $PP'$ is $-3/2$ | B1 | Could be from $-1/(2/3)$ or $(-1-5)/(5-1)$. SC: if using this method but gain 0 marks, can get B1 for sight of $P'=(5,-1)$ |
| Equation of $PP'$ is $y-5=-\frac{3}{2}(x-1)$ | M1 | Using their gradient and $(1,5)$ or their $(5,-1)$ to form the equation of the line. $y--1=-3/2(x-5)$, $y=-3/2x+13/2$ |
| $\frac{2}{3}x = -\frac{3}{2}x+\frac{13}{2} \Rightarrow x=3 \Rightarrow y=2$ | A1 [3] | |

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## Question 6(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-3/2$ | B1FT 2.2a | FT their $-1/m$ from (b). **B0** for $2/3$ |
| (Since T is a reflection) the invariant lines are the lines perpendicular to the mirror line (which reflect onto themselves) | B1 [2] 2.4 | |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{13}\begin{pmatrix}5 & 12\\12 & -5\end{pmatrix}\begin{pmatrix}x\\ax+2\end{pmatrix} = \frac{1}{13}\begin{pmatrix}5x+12ax+24\\12x-5ax-10\end{pmatrix}$ and $(12x-5ax-10)/13 = a(5x+12ax+24)/13+2$ | M1 | Multiplying any point on the line $y=ax+2$ into **A** and specifying that the image point lies on the same straight line |
| $\Rightarrow 12x-5ax-10=5ax+12a^2x+24a+26 \Rightarrow (12a^2+10a-12)x+36+24a=0$. But true for any $x \Rightarrow 36+24a=0 \Rightarrow a=-3/2$ | A1 [2] | If $12a^2+10a-12=0$ leading to $(3a-2)(2a+3)=0$ then $a=2/3$ must be properly rejected for A1 |

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