(a)
Points A(2, 20), B(14, 24)
Slope of AB: \( \frac{24 – 20}{14 – 2} = \frac{4}{12} = \frac{1}{3} \) (M1)
Perpendicular slope: \( -\frac{1}{\frac{1}{3}} = -3 \) (A1)
Midpoint of AB: \( \left( \frac{2 + 14}{2}, \frac{20 + 24}{2} \right) = (8, 22) \) (A1)
Equation: \( y = -3x + c \), substitute (8, 22): \( 22 = -3 \times 8 + c \), \( c = 22 + 24 = 46 \) (A1)
Result: \( y = -3x + 46 \) [4]

(b)
Road equation: \( -x + y = 4 \), or \( y = x + 4 \)
Perpendicular bisector: \( y = -3x + 46 \) (from (a))
Solve intersection: \( x + 4 = -3x + 46 \), \( 4x = 42 \), \( x = 10.5 \) (M1)
Substitute: \( y = 10.5 + 4 = 14.5 \) (A1)
Verify distances: \( \text{A to bus stop} = \sqrt{(10.5 – 2)^2 + (14.5 – 20)^2} = \sqrt{102.5} \), \( \text{B to bus stop} = \sqrt{(14 – 10.5)^2 + (24 – 14.5)^2} = \sqrt{102.5} \) (equal) (A1)
Result: (10.5, 14.5) [3]