a) Points: \( A(2, 20) \), \( B(14, 24) \).
Slope of \( AB \): \( \frac{24 – 20}{14 – 2} = \frac{4}{12} = \frac{1}{3} \).
Perpendicular slope: \( -\frac{1}{\frac{1}{3}} = -3 \).
Midpoint of \( AB \): \( \left( \frac{2 + 14}{2}, \frac{20 + 24}{2} \right) = (8, 22) \).
Line: \( y = -3x + c \).
Use midpoint: \( 22 = -3 \cdot 8 + c \implies 22 = -24 + c \implies c = 46 \).
Equation: \( y = -3x + 46 \). [4]

b) Road: \( -x + y = 4 \implies y = x + 4 \).
Intersect with \( y = -3x + 46 \): \( x + 4 = -3x + 46 \).
Solve: \( 4x + 4 = 46 \implies 4x = 42 \implies x = 10.5 \).
Then: \( y = 10.5 + 4 = 14.5 \).
Coordinates: \( (10.5, 14.5) \).
Verify: Distance \( A \) to bus stop: \( \sqrt{(10.5 – 2)^2 + (14.5 – 20)^2} = \sqrt{8.5^2 + (-5.5)^2} = \sqrt{102.5} \).
Distance \( B \) to bus stop: \( \sqrt{(14 – 10.5)^2 + (24 – 14.5)^2} = \sqrt{3.5^2 + 9.5^2} = \sqrt{102.5} \).
Equal distances confirm. [3]