(a)
Gradient of \( AB \): \( \dfrac{24 – 20}{14 – 2} = \dfrac{4}{12} = \dfrac{1}{3} \). M1
Midpoint of \( AB \): \( \left( \dfrac{2 + 14}{2}, \dfrac{20 + 24}{2} \right) = (8, 22) \). A1
Perpendicular slope: negative reciprocal of \( \dfrac{1}{3} = -3 \). M1
Equation using point \( (8, 22) \): \( y – 22 = -3(x – 8) \), simplify: \( y = -3x + 46 \). A1
[4 marks]
(b)
Solve \( y = -3x + 46 \) and \( y = x + 4 \): \( x + 4 = -3x + 46 \). M1
Solve: \( 4x = 42 \), \( x = \dfrac{42}{4} = 10.5 \). A1
Substitute: \( y = 10.5 + 4 = 14.5 \). Coordinates: \( (10.5, 14.5) \). A1
Verify: In \( y = -3x + 46 \), \( y = -3 \times 10.5 + 46 = 14.5 \); in \( -x + y = 4 \), \( -10.5 + 14.5 = 4 \).
[3 marks]
