Higher November 2021 Paper 3 Q16
16 \(P\) is the point (2, 14)
\(Q\) is the point (6, 8)
\(R\) is the point (2, 5)
Use gradients to show that angle \(PQR\) is not a right angle. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (Gradient of \(PQ\) =) \(\dfrac{14 - 8}{2 - 6}\) or \(\dfrac{8 - 14}{6 - 2}\) or \(-1.5\) or \(\dfrac{-3}{2}\) or (gradient of \(QR\) =) \(\dfrac{8 - 5}{6 - 2}\) or \(\dfrac{5 - 8}{2 - 6}\) or 0.75 or \(\dfrac{3}{4}\) or \(\dfrac{-3}{-4}\) | M1 | oe |
| (Gradient of \(PQ\) =) \(-1.5\) or \(\dfrac{-3}{2}\) and (gradient of \(QR\) =) 0.75 or \(\dfrac{3}{4}\) or \(\dfrac{-3}{-4}\) | M1dep | oe |
| No and \(-1.5 \times 0.75 \ne -1\) or No and \(-1.5 \times 0.75 = -1.125\) | A1ft | oe eg No and \(\dfrac{-3}{2}\) \(\times\) \(\dfrac{3}{4}\) \(= -\)\(\dfrac{9}{8}\) ft their two gradients with M1 scored accept No and \(-1.5\) is not the negative reciprocal of 0.75 |
Additional guidance
| Accept \(-\)\(\dfrac{3}{2}\) or \(\dfrac{3}{-2}\) for \(\dfrac{-3}{2}\) | |
| Gradient of \(PQ\) = \(\dfrac{-3}{2}\), gradient of \(QR\) = \(\dfrac{4}{3}\), No and \(\dfrac{-3}{2}\) \(\times\) \(\dfrac{4}{3}\) \(= -2\) | M1M0A1ft |
| Answers involving Pythagoras’ theorem or scale drawing | M0M0A0 |