FP1 June 2018 Q6
6. \[\mathbf{M} = \begin{pmatrix} 8 & -1 \\ -4 & 2 \end{pmatrix}\]
The triangle \(T\) has vertices at the points \((4, 1)\), \((6, k)\) and \((12, 1)\), where \(k\) is a constant.
The triangle \(T\) is transformed onto the triangle \(T^{\prime}\) by the transformation represented by the matrix M.
Given that the area of triangle \(T^{\prime}\) is 216 square units,
| Scheme | Marks |
|---|---|
| \(\left\{\det\mathbf{M} = (8)(2) - (-1)(-4)\right\} \Rightarrow \det\mathbf{M} = 12\) | B1 |
| (1) |
Notes
B1: 12
| Scheme | Marks |
|---|---|
| Area \(T = \dfrac{216}{12}\ \{= 18\}\) | M1 |
| \(h = \pm(1 - k)\) | M1 |
| \(\dfrac{1}{2}8(k - 1) = 18\) or \(\dfrac{1}{2}8(1 - k) = 18\) or \((k - 1) = \dfrac{18}{4}\) or \((1 - k) = \dfrac{18}{4}\) or \(\left\{\dfrac{1}{2}8h = 18\right\} \Rightarrow h = \dfrac{9}{2},\ k = 1 \pm \dfrac{9}{2}\) | ddM1 |
| \(\Rightarrow k = 5.5\) or \(k = -3.5\) | A1 A1 |
| (5) | |
| (6 marks) |
Notes
M1: Area \(T = \dfrac{216}{\text{their "}\det\mathbf{M}\text{"}}\)
M1: Uses \((k - 1)\) or \((1 - k)\) in their solution.
ddM1: dependent on the two previous M marks
\(\dfrac{1}{2}8(k - 1)\) or \(\dfrac{1}{2}8(1 - k) = \dfrac{216}{\text{their "}\det\mathbf{M}\text{"}}\)
or \((k - 1)\) or \((1 - k) = \dfrac{216}{4(\text{their "}\det\mathbf{M}\text{"})}\)
or \(h = \dfrac{216}{4(\text{their "}\det\mathbf{M}\text{"})},\ k = 1 \pm \dfrac{216}{4(\text{their "}\det\mathbf{M}\text{"})}\)
A1: At least one of either \(k = 5.5\) or \(k = -3.5\)
A1: Both \(k = 5.5\) and \(k = -3.5\)
ALT (b)
| Scheme | Marks |
|---|---|
| \(\mathbf{T}^{\prime} = \begin{pmatrix} 8 & -1 \\ -4 & 2 \end{pmatrix}\begin{pmatrix} 4 & 6 & 12 \\ 1 & k & 1 \end{pmatrix}\) | |
| \(\mathbf{T}^{\prime} = \begin{pmatrix} 31 & 48 - k & 95 \\ -14 & -24 + 2k & -46 \end{pmatrix}\) or 18 seen | M1 |
| \(\dfrac{1}{2}\begin{vmatrix} 31 & 48 - k & 95 & 31 \\ -14 & -24 + 2k & -46 & -14 \end{vmatrix} = 216\) or \(\dfrac{1}{2}\begin{vmatrix} 4 & 6 & 12 & 4 \\ 1 & k & 1 & 1 \end{vmatrix} = 18\) | M1 |
| \(\dfrac{1}{2}\left|-744 + 62k + 672 - 14k - 2208 + 46k + 2280 - 190k - 1330 + 1426\right| = 216\) \(\dfrac{1}{2}\left|4k - 6 + 6 - 12k + 12 - 4\right| = 18\) | ddM1 |
| \(\dfrac{1}{2}\left|96 - 96k\right| = 216\) or \(\dfrac{1}{2}\left|8 - 8k\right| = 18\) | |
| So, \(1 - k = 4.5\) or \(k - 1 = 4.5\) | |
| \(\Rightarrow k = -3.5\) or \(k = 5.5\) | A1 A1 |
| (5) |
M1: At least 5 out of 6 elements are correct or 18 seen..
M1: \(\dfrac{1}{2}\left|\text{their } \mathbf{T}^{\prime}\right| = 216\) or \(\dfrac{1}{2}\begin{vmatrix} 4 & 6 & 12 & 4 \\ 1 & k & 1 & 1 \end{vmatrix} = 18\)
ddM1: Dependent on the two previous M marks. Full method of evaluating a determinant.
A1: At least one of either \(k = -3.5\) or \(k = 5.5\)
A1: Both \(k = -3.5\) and \(k = 5.5\)