(a)
Perpendicular bisector of \( [BC] \) passes through \( (0, -1) \) and \( (2, 0) \)
Slope: \( \frac{0 – (-1)}{2 – 0} = \frac{1}{2} = 0.5 \) (M1)
Equation using \( (0, -1) \): \( y + 1 = 0.5x \), so \( y = 0.5x – 1 \) (A1)
Result: \( y = 0.5x – 1 \) [2]

(b)
Perpendicular bisector of \( [AB] \): \( y = -3x + 23 \)
Perpendicular bisector of \( [BC] \): \( y = 0.5x – 1 \) (from (a))
Solve intersection: \( -3x + 23 = 0.5x – 1 \), \( 24 = 3.5x \), \( x = \frac{24}{3.5} = \frac{48}{7} \approx 6.857142857 \) (M1)
Substitute \( x = \frac{48}{7} \): \( y = 0.5 \times \frac{48}{7} – 1 = \frac{24}{7} – \frac{7}{7} = \frac{17}{7} \approx 2.428571429 \)
To 4 significant figures: \( x \approx 6.857 \), \( y \approx 2.429 \) (A1)(A1)
Verify with \( y = -3x + 23 \): \( y = -3 \times 6.857 + 23 \approx 2.429 \) (A1)
Result: \( (6.857, 2.429) \) [4]

(c)
Voronoi edges are segments of perpendicular bisectors of \( [AB] \), \( [AC] \), and \( [BC] \), meeting at \( V(6.857, 2.429) \)
Draw edges from \( V \):
– Along \( y = -3x + 23 \) (for \( [AB] \))
– Along \( y = 0.5x – 1 \) (for \( [BC] \))
– Along bisector of \( [AC] \), passing through \( V \), forming cell boundaries
Diagram: (A1)(A1)
Result: Edges as shown in the diagram [2]