November 2021 Paper 2 Q5
5 \(SQR\) is a right-angled triangle.
\(P\) is a point on \(SQ\).
Angle \(SPR = 45^\circ\)
\(M\) is the midpoint of \(QR\).
\(k\) is a constant.
Not drawn accurately
Work out the coordinates of \(M\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(k - 1 = 15 - 6\) or \(\tan 45 = \dfrac{k - 1}{15 - 6}\) | M1 | oe equation eg \(k - 1 = 9\) or \(\dfrac{k - 1}{9} = 1\) |
| \((k =)\ 15 - 6 + 1\) or \((k =)\ (15 - 6)\tan 45 + 1\) or \((k =)\ 10\) | M1dep | oe eg \((k =)\ 9 + 1\) may be seen on diagram |
| (5.5, 8) | A1 | oe eg \(\left(5\dfrac{1}{2},\ 8\right)\) or \(\left(\dfrac{11}{2},\ 8\right)\) SC1 answer \((\ldots, 8)\) |
Additional guidance
| First M1 can be scored using Pythagoras’ theorem and cosine rule eg \((PR^2 =)\ 9^2 + (15 - 6)^2\) or 162 and \((PQ =)\ 6 - 1\) or 5 and (angle \(QPR\) =) \(180 - 45\) or 135 and \((k - 1)^2 + (15 - 1)^2 =\) their 162 \(+\) (their 5)\(^2\) \(-\ 2 \times \sqrt{\text{their } 162} \times\) their 5 \(\times \cos\) their 135 | M1 |
| 10 but not seen or implied to be \(k\) | M0M0 |