AS June 2022 Paper 1 Q6
6 The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \dfrac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\).
You are given that \(\mathbf{A}\) represents the transformation T which is a reflection in a certain straight line. You are also given that this straight line, the mirror line, passes through the origin, \(O\).
The coordinates of the point \(P\) are \((1, 5)\).
Determine the value of \(a\). [2]
| Scheme | Marks | AO |
|---|---|---|
| T is a reflection (in 2-D) and in any reflection any point on the mirror line remains invariant... | B1 | 2.4 |
| ...and so the mirror line must itself be a line of invariant points. | B1 | 2.2a |
| [2] |
Notes
B1: (1st) Any point on the mirror line stays where it is...
B1: (2nd) ...so the mirror line is a line of invariant points.
Accept “so the line of invariant points is the mirror line”
If B0B0 then SC1 for any answer which is, in effect, a statement that the mirror line is an invariant line.
| Scheme | Marks | AO |
|---|---|---|
| For line of invariant points \(\mathbf{Ar} = \mathbf{r}\) | B1 | 1.1 |
| \(\frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{13}\begin{pmatrix} 5x + 12y \\ 12x - 5y \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} x \\ y \end{pmatrix} \Rightarrow \dfrac{1}{13}(5x + 12y) = x\) or \(\dfrac{1}{13}(12x - 5y) = y\) | M1 | 1.1 |
| \(12y = 8x\) (or \(18y = 12x\)) \(\Rightarrow y = \dfrac{2}{3}x\) or \(y = \dfrac{2}{3}x + 0\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) Multiplying general point into \(\mathbf{A}\)
Could be awarded for sight of \(5x + 12y\) or \(12x - 5y\) o.e.
M1: (2nd) Equating and deriving an equation relating \(x\) and \(y\)
A1: Need to check that both equations give same straight line.
Alternative method
| Scheme | Marks |
|---|---|
| Line passes through \(O \Rightarrow c = 0\) | B1 |
| \(\frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix}\) | M1 |
| \(= \dfrac{1}{13}\begin{pmatrix} 5x + 12mx \\ 12x - 5mx \end{pmatrix} = \begin{pmatrix} x \\ mx \end{pmatrix}\) | M1 |
| \(5x + 12mx = 13x\) and \(12x - 5mx = 13mx \Rightarrow\) \(m = 2/3\) so \(y = \dfrac{2}{3}x\) or \(y = \dfrac{2}{3}x + 0\) | A1 |
B1: Used in the solution
M1: (1st) Considering the matrix acting on a general point on the line \(y = mx\ (+ c)\)
M1: (2nd) Multiplying and equating
A1: Need to check that both equations are satisfied by \(m = 2/3\).
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix}\) \(= \frac{1}{13}\begin{pmatrix} 5x + 12mx + 12c \\ 12x - 5mx - 5c \end{pmatrix} = \begin{pmatrix} x^{\prime} \\ y^{\prime} \end{pmatrix}\) | M1 |
| \(y' = mx' + c \Rightarrow\) \(\dfrac{1}{13}(12x - 5mx - 5c) = \dfrac{m}{13}(5x + 12mx + 12c) + c\) \(x(12m^2 + 10m - 12) + c(12m + 18) = 0\) \(2x(2m + 3)(3m - 2) + 6c(2m + 3) = 0\) | M1 |
| Either we have \(m = \frac{-3}{2}\), and \(c\) can be anything or \(m = \frac{2}{3}\) and we have \(c = 0\). | A1 |
| This gives a single line and infinitely many which are perpendicular to it. Therefore the reflection line is the single line (and the perpendicular ones are invariant lines). Hence we have \(m = \frac{2}{3}\) and so \(y = \frac{2}{3}x\) | A1 |
M1: (1st) Considering the matrix acting on a general point on the line \(y = mx + c\)
M1: (2nd) Multiplying and substituting into \(y = mx + c\)
A1: (1st) Finding two correct values of \(m\) and no others (linking to \(c\) not necessary here)
A1: (2nd) Convincing reason why \(m = -3/2\) is rejected as a possibility.
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}\) [so \(P^{\prime}\) is \((5, -1)\)] | M1 | 1.1 |
| So required \(x\)-coord is \(\tfrac{1}{2}(5 + 1) = 3\ldots\) | A1 | 2.2a |
| ...and required \(y\)-coord is \(\tfrac{1}{2}(5 + -1) = 2\) | A1 | 1.1 |
| [3] |
Notes
SC: If answer given as a vector then maximum mark is M1 A1 A0
M1 could be awarded for solving \(\mathbf{Ax} = (5, 1)\) to get \(\mathbf{x} = (5, -1)\)
Alternative method
| Scheme | Marks |
|---|---|
| Gradient of line \(PP'\) is \(-3/2\) | B1 |
| Equation of \(PP'\) is \(y - 5 = -3/2\,(x - 1)\) | M1 |
| \(2/3\,x = -3/2\,x + 13/2 \Rightarrow x = 3 \Rightarrow y = 2\) | A1 |
B1: Could be from \(-1/(2/3)\) or \((-1 - 5)/(5 - 1)\)
M1: Using their gradient and \((1, 5)\) or their \((5, -1)\) to form the equation of the line
\(y - -1 = -3/2\,(x - 5)\)
\(y = -3/2\,x + 13/2\)
SC: if using this method but gain 0 marks, can get B1 for sight of \(P' = (5, -1)\)
| Scheme | Marks | AO |
|---|---|---|
| \(-3/2\) | B1FT | 2.2a |
| (Since T is a reflection) the invariant lines are the lines perpendicular to the mirror line (which reflect onto themselves) | B1 | 2.4 |
| [2] |
Notes
B1FT: FT their \(-1/m\) from (b)
B0 for 2/3.
Alternative method
| Scheme | Marks |
|---|---|
| \(\frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}\begin{pmatrix} x \\ ax + 2 \end{pmatrix} = \frac{1}{13}\begin{pmatrix} 5x + 12ax + 24 \\ 12x - 5ax - 10 \end{pmatrix}\) and \((12x - 5ax - 10)/13 = a(5x + 12ax + 24)/13 + 2\) | M1 |
| \(\Rightarrow 12x - 5ax - 10 = 5ax + 12a^2x + 24a + 26\) \(\Rightarrow (12a^2 + 10a - 12)x + 36 + 24a = 0\) But true for any \(x \Rightarrow 36 + 24a = 0 \Rightarrow a = -3/2\) | A1 |
M1: Multiplying any point on the line \(y = ax + 2\) into \(\mathbf{A}\) and specifying that the image point lies on the same straight line.
A1: If \(12a^2 + 10a - 12 = 0\) leading to \((3a - 2)(2a + 3) = 0\) then \(a = 2/3\) must be properly rejected for A1.