A2 June 2024 Paper 2 Q12
12 The transformation S is represented by the matrix \(\mathbf{M} = \begin{bmatrix} 1 & -6 \\ 2 & 7 \end{bmatrix}\)
The transformation T is a reflection in the line \(y = x\sqrt{3}\) and is represented by the matrix \(\mathbf{N}\)
The point \(P(x, y)\) is transformed first by S, then by T
The result of these transformations is the point \(Q(3, 8)\)
Find the coordinates of \(P\)
Give your answers to three decimal places. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\begin{bmatrix} -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\[1ex] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\) | B1 | 2.2a |
| Operates first \(\mathbf{M}\), then their \(\mathbf{N}\), on column vector \(\begin{bmatrix} x \\ y \end{bmatrix}\) or Operates their \(\mathbf{N}\) or their \(\mathbf{N}^{-1}\) on \(\begin{bmatrix} 3 \\ 8 \end{bmatrix}\) or Calculates matrix product \(\mathbf{NM}\) | M1 | 3.1a |
| Obtains a correct system of simultaneous equations in \(x\) and \(y\) or Obtains correct value of \(\mathbf{M}^{-1}\) (Accept decimal approximation.) or Obtains correct value of \(\mathbf{NM}\) PI | A1 | 1.1b |
| Solves their simultaneous equations in \(x\) and \(y\) from \(\mathbf{NM}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 8 \end{bmatrix}\) or \(\mathbf{MN}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 8 \end{bmatrix}\) or Operates \(\mathbf{M}^{-1}\) on \(\mathbf{N}Q\) or Uses inverse of \(\mathbf{NM}\), where \((\mathbf{NM})^{-1} = \begin{bmatrix} 0.08927 & 0.47696 \\ 0.09821 & -0.06484 \end{bmatrix}\) to obtain coordinates of \(P\) Condone applying transformations in the wrong order. | M1 | 1.1a |
| Obtains AWRT 4.083 and AWRT \(-0.224\). Condone column vector form if \(x\) and \(y\) seen. Accept exact values. \(x = \dfrac{27 + 74\sqrt{3}}{38}\), \(y = \dfrac{14 - 13\sqrt{3}}{38}\) | A1 | 1.1b |
| (5 marks) |
Typical solution
T is a reflection in the line \(y = x\tan\dfrac{\pi}{3}\)
\[\text{So } \mathbf{N} = \begin{bmatrix} -\dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\[1ex] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{bmatrix}\]\[\mathbf{M}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -6 \\ 2 & 7 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x - 6y \\ 2x + 7y \end{bmatrix}\]\[\begin{bmatrix} 3 \\ 8 \end{bmatrix} = \mathbf{N}\left(\mathbf{M}\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} -\dfrac{1}{2}(x - 6y) + \dfrac{\sqrt{3}}{2}(2x + 7y) \\[1ex] \dfrac{\sqrt{3}}{2}(x - 6y) + \dfrac{1}{2}(2x + 7y) \end{bmatrix}\]\[\begin{bmatrix} 3 \\ 8 \end{bmatrix} = \begin{bmatrix} \left(\sqrt{3} - \dfrac{1}{2}\right)x + \left(3 + \dfrac{7\sqrt{3}}{2}\right)y \\[1ex] \left(\dfrac{\sqrt{3}}{2} + 1\right)x + \left(\dfrac{7}{2} - 3\sqrt{3}\right)y \end{bmatrix}\]\[1.23205x + 9.06218y = 3\]\[1.86603x - 1.69615y = 8\]\[x = 4.083,\ y = -0.224\]\[P(4.083, -0.224)\](3 d.p.)