(a)
Apply transformation: \( \begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} 7 & -10 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} 6 \\ -2 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} \) (M1)
Matrix multiplication: \( \begin{pmatrix} 7 \times 6 + (-10) \times (-2) \\ 2 \times 6 + (-3) \times (-2) \end{pmatrix} = \begin{pmatrix} 42 + 20 \\ 12 + 6 \end{pmatrix} = \begin{pmatrix} 62 \\ 18 \end{pmatrix} \)
Add translation: \( \begin{pmatrix} 62 \\ 18 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 62 – 5 \\ 18 + 4 \end{pmatrix} = \begin{pmatrix} 57 \\ 22 \end{pmatrix} \) (A1)
Result: (57, 22) [2]

(b)
Set up: \( \begin{pmatrix} 7 & -10 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 2p \\ 2q \end{pmatrix} \) (M1)
Equations: \( 7p – 10q – 5 = 2p \), \( 2p – 3q + 4 = 2q \)
Simplify: \( 5p – 10q = 5 \), \( 2p – 5q = -4 \) (A1)
Solve: Multiply first by 2: \( 10p – 20q = 10 \), subtract second: \( (10p – 20q) – (2p – 5q) = 10 – (-4) \), \( 8p – 15q = 14 \)
Solve with first: \( 5p – 10q = 5 \), multiply by 3: \( 15p – 30q = 15 \), multiply second by 2: \( 16p – 30q = 28 \), subtract: \( p = 13 \)
Substitute: \( 5 \times 13 – 10q = 5 \), \( 65 – 10q = 5 \), \( q = 6 \) (A1)
Verify: \( 7 \times 13 – 10 \times 6 – 5 = 26 = 2 \times 13 \), \( 2 \times 13 – 3 \times 6 + 4 = 12 = 2 \times 6 \)
Result: \( p = 13 \), \( q = 6 \) [3]

(c)
Linear part: \( A = \begin{pmatrix} 7 & -10 \\ 2 & -3 \end{pmatrix} \), compute determinant: \( \det A = (7 \times (-3)) – ((-10) \times 2) = -21 – (-20) = -1 \) (M1)
Area scale factor: \( |\det A| = |-1| = 1 \)
Translation \( \begin{pmatrix} -5 \\ 4 \end{pmatrix} \) does not affect area (A1)
Explanation: Area unchanged as \( |\det A| = 1 \), and translation preserves area
Result: Triangle \( L \) and its image have the same area [2]