FP1 June 2014 Q7
7.
(i) In each of the following cases, find a \(2 \times 2\) matrix that represents
(a) a reflection in the line \(y = -x\),
(b) a rotation of \(135^\circ\) anticlockwise about \((0, 0)\),
(c) a reflection in the line \(y = -x\) followed by a rotation of \(135^\circ\) anticlockwise about \((0, 0)\). (4)
(ii) The triangle \(T\) has vertices at the points \((1, k)\), \((3, 0)\) and \((11, 0)\), where \(k\) is a constant.
Triangle \(T\) is transformed onto the triangle \(T^{\prime}\) by the matrix \[\begin{pmatrix} 6 & -2 \\ 1 & 2 \end{pmatrix}\]
Given that the area of triangle \(T^{\prime}\) is 364 square units, find the value of \(k\). (6)| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) | B1 |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\) | B1 |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\) M1: Multiplies their (b) x their (a) in the correct order A1: Correct matrix Correct matrix seen M1A1 | M1A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Area triangle T \(= \dfrac{1}{2} \times (11 - 3) \times k = 4k\) M1: Correct method for area for \(T\) A1: 4k | M1A1 |
| \(\det\begin{pmatrix} 6 & -2 \\ 1 & 2 \end{pmatrix} = 6 \times 2 - 1 \times (-2)\ (= 14)\) M1: Correct method for determinant A1: 14 | M1A1 |
| Area triangle \(T = \dfrac{364}{\text{"}14\text{"}}\ (= 26) \Rightarrow 4k = 26\) Uses 364 and their determinant correctly to form an equation in \(k\). | M1 |
| \(k = \dfrac{26}{4}\ \left(= \dfrac{13}{2}\right)\) Accept \(k = \pm\dfrac{13}{2}\) or \(k = -\dfrac{13}{2}\) | A1 |
| (6) | |
| (10 marks) |