Home / IB Mathematics SL 3.5 Equations of perpendicular bisectors AI SL Paper 2 – Exam Style Questions

IB Mathematics SL 3.5 Equations of perpendicular bisectors AI SL Paper 2 - Exam Style Questions - New Syllabus

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Question

The map below illustrates the positions of three fire stations—\( A \), \( B \), and \( C \)—serving a \( 50\text{ km} \) by \( 50\text{ km} \) rural district in Japan. To manage emergency responses efficiently, a Voronoi diagram is employed to partition the district into regions, ensuring each location is assigned to the nearest station. The corresponding Voronoi diagram is plotted on a coordinate grid where \( 1 \) unit represents \( 2.5\text{ km} \).
Voronoi diagram of fire stations A, B, and C
The approximate coordinates for the stations are given as \( A(8, 13) \), \( B(16, 12) \), and \( C(8, 13) \).
(a) (i) Calculate the coordinates of the midpoint of the line segment \( [BC] \).
(ii) Determine the gradient of the perpendicular bisector of \( [BC] \).
(iii) Find the equation of the Voronoi edge separating the service areas of stations \( B \) and \( C \).
(b) (i) Based on the Voronoi model, which fire station is responsible for a fire reported at coordinates \( (14, 10) \)? Give a reason for your choice.
(ii) Provide one practical reason why a different station might actually be dispatched to this fire.
An emergency is reported at location \( D \) with coordinates \( \left( 11\frac{1}{3}, 7\frac{1}{6} \right) \). The calculated distance from \( D \) to station \( A \) on the grid is \( 6.71855 \) units (\( 16.7964\text{ km} \)).
(c) (i) Verify that the distance from \( D \) to station \( B \) is also exactly \( 6.71855 \) units.
(ii) Explain why any of the three fire stations could be considered equally responsible for responding to the fire at location \( D \).
The boundary between regions \( A \) and \( B \) is defined by the line \( y = 0.5x + 1.5 \). Furthermore, the boundary between regions \( A \) and \( C \) has an \( x \)-intercept at \( 7.75 \).
(d) (i) Calculate the area of the Voronoi cell for which fire station \( A \) is responsible, rounding to the nearest square unit.
(ii) Determine the actual geographic area, in square kilometers, served by station \( A \).

Most-appropriate topic codes (IB Mathematics AI SL/HL 2025):

• SL 3.1: Geometry of 2D shapes and distance formula — part (c)
• SL 3.5: Equations of perpendicular bisectors — part (a)
• SL 3.6: Voronoi diagrams: sites, cells, and edges — parts (b), (c)
▶️ Answer/Explanation

(a)
(i) B(16, 12), C(8, 13). Midpoint = \( \left( \dfrac{16+8}{2}, \dfrac{12+13}{2} \right) = (12, 12.5) \).
\( \boxed{(12, 12.5)} \)

(ii) Gradient of BC = \( \dfrac{13-12}{8-16} = \dfrac{1}{-8} = -0.125 \).
Gradient of perpendicular bisector = \( 8 \).
\( \boxed{8} \)

(iii) Using point (12, 12.5) and gradient 8:
\( y – 12.5 = 8(x – 12) \)
\( y = 8x – 83.5 \).
\( \boxed{y = 8x – 83.5} \)

(b)
(i) Point (14, 10) lies in the cell of fire station B in the Voronoi diagram. Therefore B is closest and should respond.
The point is in B’s Voronoi cell / closest to B.

(ii) Possible reasons:
• B may be busy with another incident.
• Road access might be better from another station.
• Another station may have specialized equipment.
Any plausible practical reason.

(c)
(i) D \( \left( 11\frac{1}{3}, 7\frac{1}{6} \right) \), B(16, 12).
\( DB = \sqrt{(16 – 11\frac{1}{3})^2 + (12 – 7\frac{1}{6})^2} \)
= \( \sqrt{(4\frac{2}{3})^2 + (4\frac{5}{6})^2} \)
= \( \sqrt{\frac{196}{9} + \frac{841}{36}} \)
= \( \sqrt{\frac{784+841}{36}} = \sqrt{\frac{1625}{36}} \approx 6.71855 \).

(ii) Similarly, DC ≈ 6.71855. Since D is equidistant from A, B, and C, it lies at the intersection of Voronoi edges. Any station could respond.

(d)
(i) Vertices of A’s cell: intersection of \( y = 8x – 83.5 \) and \( y = 0.5x + 1.5 \) gives D \( (11\frac{1}{3}, 7\frac{1}{6}) \).
Intersection of \( y = 0.5x + 1.5 \) and x = 7.75 gives point (7.75, 5.375).
Area can be found by integration or geometry. Using triangles/trapezoids:
Area ≈ 93.73 units² → 94 units² (nearest unit).
\( \boxed{94 \text{ units}^2} \)

(ii) 1 unit = 2.5 km → 1 unit² = 6.25 km².
Actual area = \( 94 \times 6.25 = 587.5 \text{ km}^2 \).
\( \boxed{588 \text{ km}^2} \) (nearest km²).

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