# Question 2:

## Part (a) — Way 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{grad } PQ = \frac{11-7}{9--3} = \frac{1}{3}$, $\text{grad } QR = \frac{11-2}{9-12} = -3$ | M1 A1 | M1: Attempts both gradients with $\frac{\text{difference in }y}{\text{difference in }x}$ seen at least once. A1: Both correct gradients (may be unsimplified e.g. $\frac{4}{12}$ and $\frac{9}{-3}$). |
| $\frac{1}{3} \times -3 = -1$ so angle $PQR = 90°$ | A1 | A1: e.g. $\frac{1}{3} \times -3 = -1 \Rightarrow PQR = 90°$. Must include explanation and conclusion referencing angle $PQR = 90°$ or equivalent. Do not allow ambiguous statements. |

## Part (a) — Way 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $PQ^2 = (9--3)^2 + (11-7)^2 = 160$, $QR^2 = (12-9)^2 + (2-11)^2 = 90$, $PR^2 = (12--3)^2 + (2-7)^2 = 250$ | M1 A1 | M1: Attempts all three lengths with "difference of coordinates" and squaring seen at least twice. A1: All three lengths or lengths² correct as exact terms or decimals (1 d.p.). |
| $PQ^2 + QR^2 = PR^2$ (i.e. $90 + 160 = 250$) so angle $PQR = 90°$, or e.g. $\cos\theta = \frac{160+90-250}{2\sqrt{160}\sqrt{90}} = 0 \Rightarrow \theta = 90°$ | A1 | A1: Via Pythagoras or cosine rule with correct conclusion. Values must be exact for this mark. |

## Part (a) — Way 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PQ} = \begin{pmatrix}12\\4\end{pmatrix}$, $\overrightarrow{QR} = \begin{pmatrix}3\\-9\end{pmatrix}$ | M1 A1 | M1: Attempts $\pm\overrightarrow{PQ}$ and $\pm\overrightarrow{QR}$ with "difference of coordinates" seen at least twice. A1: Correct vectors for $\pm\overrightarrow{PQ}$ and $\pm\overrightarrow{QR}$. |
| $\overrightarrow{PQ} \cdot \overrightarrow{QR} = \begin{pmatrix}12\\4\end{pmatrix} \cdot \begin{pmatrix}3\\-9\end{pmatrix} = 36 - 36 = 0$ so angle $PQR = 90°$ | A1 | A1: $\overrightarrow{PQ}\cdot\overrightarrow{QR} = 0$ so angle $PQR = 90°$. |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $(-3,7)+(3,-9) = \ldots$ or $(12,2)-(12,4) = \ldots$ | M1 | M1: Any suitable method of finding at least $x$ or $y$ for $S$; can be implied by one correct coordinate. |
| $(0, -2)$ | A1 | A1: Correct coordinates $(0,-2)$, may be written as $x=0,\ y=-2$. |

**ALT 1(b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{grad }PQ = \frac{1}{3} \Rightarrow$ eqn $RS$: $y-2=\frac{1}{3}(x-12)$; $\text{grad }QR=-3 \Rightarrow$ eqn $PS$: $y-7=-3(x+3)$; solve simultaneously | M1 | M1: Attempts equations of lines $PS$ and $RS$, solves simultaneously for $x$ or $y$. |
| $(0,-2)$ | A1 | |

**ALT 2(b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Midpoint $PR$ is $\left(\frac{9}{2},\frac{9}{2}\right) \Rightarrow \frac{9+x}{2}=\frac{9}{2},\ \frac{11+y}{2}=\frac{9}{2}$ | M1 | M1: Attempts midpoint of $PR$ and uses $Q$ to find $S$ for at least one of $x$ or $y$. |
| $(0,-2)$ | A1 | |

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