FP1 June 2015 Q7
7.
A triangle \(T\) is transformed onto a triangle \(T^{\prime}\) by the transformation represented by the matrix \(\mathbf{B}\).
The vertices of triangle \(T^{\prime}\) have coordinates \((0, 0)\), \((-20, 6)\) and \((10c, 6c)\), where \(c\) is a positive constant.
The area of triangle \(T^{\prime}\) is 135 square units.
| Scheme | Marks |
|---|---|
| \(5k(k+1) - -3(3k-1) = 0\) | M1 |
| \(5k^2 + 5k + 9k - 3 = 0\) | A1 |
| \((5k - 1)(k + 3) = 0\) so \(k =\) | M1 |
| \(k = \dfrac{1}{5}\) or \(-3\) | A1 |
| (4) |
Notes
M1: Puts determinant equal to zero
A1: cao as three or four term quadratic
M1: Solve their quadratic to find \(k\)
A1: cao – need both correct answers
| Scheme | Marks |
|---|---|
| \(\mathbf{B}^{-1} = \dfrac{1}{45}\begin{pmatrix} 3 & -5 \\ 3 & 10 \end{pmatrix}\) | M1 A1 |
| (2) |
Notes
M1 Uses correct method for inverse with fraction \(\dfrac{1}{45}\) or \(\dfrac{1}{\text{their det}}\)
A1: All correct oe
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{45}\begin{pmatrix} 3 & -5 \\ 3 & 10 \end{pmatrix}\begin{pmatrix} 0 & -20 & 10c \\ 0 & 6 & 6c \end{pmatrix} =\) | M1 |
| \(\dfrac{1}{45}\begin{pmatrix} 0 & -90 & 0 \\ 0 & 0 & 90c \end{pmatrix}\) Vertices at \((0, 0)\) \((-2, 0)\) \((0, 2c)\) | A1,A1 |
| ALT | |
| \(\begin{pmatrix} 10 & 5 \\ -3 & 3 \end{pmatrix}\begin{pmatrix} a & d & f \\ b & e & g \end{pmatrix} = \begin{pmatrix} 0 & -20 & 10c \\ 0 & 6 & 6c \end{pmatrix}\) and attempt to form simultaneous equations | M1 |
| \(10a + 5b = 0, -3a + 3b = 0\) \(10d + 5e = -20, -3d + 3e = 6\) \(10f + 5g = 10c, -3f + 3g = 6c\) all correct oe | A1 |
| Vertices at \((0, 0)\) \((-2, 0)\) \((0, 2c)\) | A1 |
| (3) |
Notes
M1: Post multiplies their inverse by 2 by 3 matrix or 2 by 2 matrix excluding the origin or does not use inverse and attempts to form simultaneous equations. Can exclude origin.
A1: \((-2,0)\) and \((0,2c)\). Can be written as column vectors. Accept seen in final two columns of single matrix
A1: \((0,0)\). Can be written as column vectors. Award if seen as first column of single matrix.
| Scheme | Marks |
|---|---|
| Area of \(T\) is \(\tfrac{1}{2} \times 2 \times 2c = 2c\) | B1 |
| Area of \(T \times \text{determinant} = 135\) | M1 |
| So \(c = \dfrac{3}{2}\) | A1 |
| OR Area of \(T^{\prime} = \tfrac{1}{2}\begin{vmatrix} 0 & -20 & 10c & 0 \\ 0 & 6 & 6c & 0 \end{vmatrix} = 90c\) Their area = 135 So \(c = \dfrac{3}{2}\) | B1 M1 A1 |
| (3) | |
| (12 marks) |
Notes
B1: Area of \(T\) given as \(2c\) or area of \(T^{\prime} = 90c\) Accept \(\pm\)
M1: Either method using their area of \(T\) and their det or their area of \(T^{\prime}\)
A1: \(c = \dfrac{3}{2}\) cao