Higher November 2023 Paper 1 Q28
28 \(PQRS\) is a quadrilateral.
\(PQ\) is not parallel to \(SR\).
\(X\) is a point on \(QR\).
\(QX : XR = 2 : 3\)
\(\overrightarrow{QX} = 2\mathbf{a} + 4\mathbf{b}\)
Not drawn accurately
Prove that \(PQRS\) is a trapezium. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((\overrightarrow{XR} =)\ \dfrac{3}{2}(2\mathbf{a} + 4\mathbf{b})\) or \(3\mathbf{a} + 6\mathbf{b}\) or \((\overrightarrow{QR} =)\ \dfrac{5}{2}(2\mathbf{a} + 4\mathbf{b})\) or \(5\mathbf{a} + 10\mathbf{b}\) | M1 | oe may be on diagram |
| \((\overrightarrow{PS} =)\ -5\mathbf{a} + 2\mathbf{a} + 4\mathbf{b} + \dfrac{3}{2}(2\mathbf{a} + 4\mathbf{b}) + \mathbf{a} - 8\mathbf{b}\) or \((\overrightarrow{PS} =)\ -5\mathbf{a} + \dfrac{5}{2}(2\mathbf{a} + 4\mathbf{b}) + \mathbf{a} - 8\mathbf{b}\) or \((\overrightarrow{PS} =)\ \mathbf{a} + 2\mathbf{b}\) | M1dep | oe may be on diagram |
| \((\overrightarrow{PS} =)\ \mathbf{a} + 2\mathbf{b}\) and indication why \(PS\) is parallel to \(QR\) | A1 | eg \(2(\mathbf{a} + 2\mathbf{b}) = 2\mathbf{a} + 4\mathbf{b}\) or \(5\mathbf{a} + 10\mathbf{b} = 5(\mathbf{a} + 2\mathbf{b})\) or \(\mathbf{a} + 2\mathbf{b}\) and \(\overrightarrow{QR}\) is a multiple of \(\overrightarrow{PS}\) |
Additional guidance
| Some or all vectors may be reversed and the final mark can be from using a negative constant eg \(\overrightarrow{RX} = -3\mathbf{a} - 6\mathbf{b}\) \(\overrightarrow{PS} = \mathbf{a} + 2\mathbf{b}\) \(-3(\mathbf{a} + 2\mathbf{b}) = -3\mathbf{a} - 6\mathbf{b}\) | M1 M1 A1 |