AS June 2018 Paper 1 Q6
6 Find the invariant line of the transformation of the \(x\)-\(y\) plane represented by the matrix \(\begin{pmatrix} 2 & 0 \\ 4 & -1 \end{pmatrix}\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 2 & 0 \\ 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}\) \(\Rightarrow x' = 2x,\ y' = 4x - y\) | M1 | 2.1 |
| \(y = mx + c\) and \(y' = mx' + c\) | M1 | 1.1b |
| \(\Rightarrow 4x - mx - c = m.2x + c\) \(\Rightarrow m = \tfrac{4}{3},\ c = 0\) | A1 | 2.4 |
| so invariant line is \(y = \tfrac{4}{3}x\) | A1 | 1.1b |
| [4] |
Notes
M1: (1st) soi [condone \(x = 2x\), \(y = 4x - y\)]
M1: (2nd) condition for invariance
A1: (1st) forming identity in \(x\)
(may assume \(c = 0\))
assuming \(c = 0\):
\(y = mx\) and \(y' = mx'\)
\(4x - mx = m.2x\)
\(m = \tfrac{4}{3}\)