June 2025 Paper 1 Q10
10 \(PQ\) is perpendicular to \(QR\).
Work out the value of \(a\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: using gradients | ||
| (Gradient \(QR =\)) \(\dfrac{12 - 0}{3 - 9}\) or \(-2\) | M1 | oe condone embedded in an equation |
| \(\dfrac{-1}{\text{their } -2}\) or \(\dfrac{1}{2}\) | M1 | oe condone embedded in an equation if no working for first M mark \(\dfrac{1}{2}\) would imply M2 |
| \(-24 + 2a = -3\) or \(\dfrac{12 - a}{3 - 0} = \dfrac{1}{2}\) | M1dep | oe implies M3 and must be fully correct condone \(c\) instead of \(a\) |
| \(10\dfrac{1}{2}\) or \(\dfrac{21}{2}\) or 10.5 | A1 | |
| Alternative method 2: using Pythagoras | ||
| \((QR =)\ \sqrt{6^2 + 12^2}\) or \((QR^2 =)\ 6^2 + 12^2\) or \((PQ =)\ \sqrt{3^2 + (12 - a)^2}\) or \((PQ^2 =)\ 3^2 + (12 - a)^2\) or \((PR =)\ \sqrt{9^2 + a^2}\) or \((PR^2 =)\ 9^2 + a^2\) | M1 | could be seen on diagram |
| \((QR =)\ \sqrt{6^2 + 12^2}\) or \((QR^2 =)\ 6^2 + 12^2\) and \((PQ =)\ \sqrt{3^2 + (12 - a)^2}\) or \((PQ^2 =)\ 3^2 + (12 - a)^2\) and \((PR =)\ \sqrt{9^2 + a^2}\) or \((PR^2 =)\ 9^2 + a^2\) | M1 | |
| \(81 + a^2 = 9 + (12 - a)^2 + 180\) | M1dep | oe dep on M2 must be a correct equation |
| \(10\dfrac{1}{2}\) or \(\dfrac{21}{2}\) or 10.5 | A1 | |
Additional guidance
| Condone further incorrect simplification of fraction on answer line | |
| Common incorrect gradients of QR in Alt 1 will give answers of: 13.5 from gradient of \(QR = 2\) 18 from gradient of \(QR = \dfrac{1}{2}\) 6 from gradient of \(QR = -\dfrac{1}{2}\) These could be M2M0A0 or M0M1M0A0 depending on whether the first mark can be awarded |