IB DP Mathematics: Applications and Interpretation – Perpendicular bisectors and Voronoi diagrams Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers perpendicular bisectors and Voronoi diagrams in IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 3.5 and 3.6, which are common content for SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

The guide lists technology as required on every AI paper, so every question is labelled “(calculator allowed)”. Give exact answers where neat, otherwise 3 significant figures. Units are km.

Learn the methods in the study guide and the revision notes. The IB DP Maths AI course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed) Find the equation of the perpendicular bisector of P(−2, 3) and Q(4, 7). Give your answer in the form ax + by + d = 0, where a, b, d ∈ ℤ. [3]

2. (calculator allowed) A line segment lies on the line 2x − 5y + 11 = 0 and has midpoint M(2, 3).

(a) Show that M lies on the line. [1] (b) Find the equation of the perpendicular bisector of the segment in the form ax + by + d = 0, where a, b, d ∈ ℤ. [3]

3. (calculator allowed) The point R(k, 1) is the same distance from A(1, 6) as from B(7, 2). By finding the perpendicular bisector of AB, find k. [3]

4. (calculator allowed) Two phone masts are at F(2, 5) and G(8, 1). The edge between their cells lies on 3x − 2y = 9. Determine which mast is closer to the house H(6, 5). [3]

5. (calculator allowed) Four weather stations recorded this rainfall: P(1, 2) 42 mm, Q(7, 1) 35 mm, R(3, 8) 51 mm and S(9, 7) 28 mm. A farm is at F(5, 5).

(a) Find the distance from F to each station and state which is nearest. [2] (b) Use nearest neighbour interpolation to estimate the rainfall at the farm. [1] (c) State one limitation of this estimate. [1]

6. (calculator allowed) Sites are at A(1, 1), B(9, 1) and C(7, 7).

(a) Write down the equation of the perpendicular bisector of AB. [1] (b) Find the equation of the perpendicular bisector of BC. [3] (c) Hence find the coordinates of the vertex of the Voronoi diagram. [1]

7. (calculator allowed) Three sites are at J(1, 2), K(6, 4) and L(2, 7).

(a) Find the equation of the perpendicular bisector of JL in the form y = mx + c. [3] (b) The perpendicular bisector of JK is y = −2.5x + 11.75. Find the coordinates of the vertex of the Voronoi diagram. [1] (c) Find the distance from the vertex to J. [2]

8. (calculator allowed) Two villages, U and W, lie in a county modelled by 0 ≤ x ≤ 10, 0 ≤ y ≤ 10. The straight road UW lies on the line x + 2y = 14, and its midpoint is M(4, 5).

(a) Find the equation of the boundary between the cells of U and W. [3] (b) Find the coordinates of the points where this boundary meets the lines y = 0 and y = 10. [2] (c) Village U is at (0, 7). Find the area of U’s cell. [2]

9. (calculator allowed) Four towns are at A(3, 1), B(3, 5), C(7, 1) and D(11, 5) in a region 0 ≤ x ≤ 12, 0 ≤ y ≤ 10. A waste plant must go at a vertex of their Voronoi diagram, as far as possible from the nearest town.

(a) Show that the perpendicular bisector of BC has equation y = x − 2. [3] (b) The edge between B and D lies on x = 7 and the edge between C and D lies on x + y = 12. Find the vertex V where the cells of B, C and D meet, and verify that it lies on the edge between C and D. [2] (c) The other vertex is U(5, 3), where the cells of A, B and C meet. Decide where the plant should go, and state its distance from the nearest town. [3] (d) A farmhouse at (9, 8) is served by its nearest town. State which town. [1]

10. (calculator allowed) Fire stations are at A(2, 9), B(2, 7) and C(6, 7) in a district 0 ≤ x ≤ 12, 0 ≤ y ≤ 10. Their Voronoi diagram has edges on y = 8 (A and B), x = 4 (B and C) and y = 2x (A and C), meeting at (4, 8). A new station D(6, 3) is built.

(a) Show that D lies in the cell of C. [2] (b) Find the equation of the perpendicular bisector of CD. [2] (c) Find the equation of the perpendicular bisector of BD, and show that it passes through the point where your answer to (b) meets x = 4. [3] (d) State which part of the original edge x = 4 is removed. [1] (e) D’s new cell is bounded by y = 5, y = x + 1 and the edges of the district. Find its area. [3]

Answers

1. Midpoint = ((−2 + 4)/2, (3 + 7)/2) = (1, 5) [1]. Gradient PQ = 4/6 = 2/3, so the perpendicular gradient is −3/2: y − 5 = −(3/2)(x − 1) [1]. So 2y − 10 = −3x + 3, giving 3x + 2y − 13 = 0 [1] Examiner insight: The accuracy mark needs integer coefficients as asked; y = −1.5x + 6.5 is correct but not in the required form.

2. (a) 2(2) − 5(3) + 11 = 4 − 15 + 11 = 0, so M lies on the line [1] (b) 5y = 2x + 11, so the segment’s gradient is 2/5 [1]. The perpendicular gradient is −5/2: y − 3 = −(5/2)(x − 2) [1]. So 2y − 6 = −5x + 10, giving 5x + 2y − 16 = 0 [1] Examiner insight: “Show that” in (a) needs the substitution written out and the value 0 stated, not just “true”.

3. Midpoint of AB = (4, 4) and gradient AB = −4/6 = −2/3, so the perpendicular gradient is 3/2 [1]. Bisector: y − 4 = (3/2)(x − 4), so y = 1.5x − 2 [1]. Put y = 1: 1 = 1.5k − 2, so k = 2 [1] Examiner insight: The question says “by finding the perpendicular bisector”, so a distance-only method risks losing the method marks.

4. At H: 3(6) − 2(5) = 8, which is less than 9 [1]. At F: 3(2) − 2(5) = −4, also less than 9, so H is on the same side of the edge as F [1]. So H is closer to mast F [1] Examiner insight: The final mark needs a comparison with a site, or both distances; finding 8 < 9 alone does not say which mast.

5. (a) FP = 5, FQ = √20 = 4.47, FR = √13 = 3.61, FS = √20 = 4.47 [1]. R is nearest [1] (b) 51 mm [1] (c) For example: it ignores the other stations, so the estimate jumps at a cell edge [1] Examiner insight: In (b) the answer is the site’s value itself; averaging the stations is a different method and scores nothing.

6. (a) x = 5 [1] (b) Midpoint of BC = (8, 4) [1]. Gradient BC = 6/(−2) = −3, so the perpendicular gradient is 1/3 [1]. y − 4 = (1/3)(x − 8), so y = (1/3)x + 4/3 (or x − 3y + 4 = 0) [1] (c) Substitute x = 5: y = 5/3 + 4/3 = 3, so (5, 3) [1] Examiner insight: “Hence” means use (a) and (b); a wrong (b) can still earn the (c) mark by follow-through.

7. (a) Midpoint of JL = (1.5, 4.5) [1]. Gradient JL = 5/1 = 5, so the perpendicular gradient is −1/5 [1]. y − 4.5 = −0.2(x − 1.5), so y = −0.2x + 4.8 [1] (b) Solving y = −2.5x + 11.75 and y = −0.2x + 4.8 on a GDC gives (3.02, 4.20) [1] (c) Using the unrounded vertex (3.0217…, 4.1956…): d = √(2.0217…² + 2.1956…²) [1] = 2.98 km [1] Examiner insight: Using the rounded vertex (3.02, 4.20) gives 2.99, which loses the accuracy mark; keep calculator values until the end.

8. (a) From x + 2y = 14, the road’s gradient is −1/2, so the perpendicular gradient is 2 [1]. y − 5 = 2(x − 4) [1], so y = 2x − 3 [1] (b) y = 0 gives x = 1.5, so (1.5, 0) [1]. y = 10 gives x = 6.5, so (6.5, 10) [1] (c) U’s cell is the trapezium with parallel sides 1.5 and 6.5 and height 10: area = (1/2)(1.5 + 6.5)(10) [1] = 40 km² [1] Examiner insight: Using the road’s gradient −1/2 gives y = −0.5x + 7, which is the road itself and earns no accuracy mark.

9. (a) Midpoint of BC = (5, 3) [1]. Gradient BC = (1 − 5)/(7 − 3) = −1, so the perpendicular gradient is 1 [1]. y − 3 = 1(x − 5), so y = x − 2 [1] (b) Solving y = x − 2 with x = 7 gives V(7, 5) [1]. Check: 7 + 5 = 12, so V lies on x + y = 12 [1] (c) U is √((5 − 3)² + (3 − 1)²) = √8 = 2.83 km from A, B and C [1]. V is √((7 − 3)² + (5 − 5)²) = 4 km from B, C and D [1]. The larger distance is at V, so place the plant at V(7, 5), 4 km from the nearest towns [1] (d) Town D (√13 = 3.61 km) [1] Examiner insight: In (c) you must compare a distance at each vertex; choosing V without calculating the distance at U does not earn the final mark.

10. (a) DA = √52 = 7.21, DB = √32 = 5.66, DC = 4 [1]. DC is the smallest, so D lies in C’s cell [1] (b) CD is vertical with midpoint (6, 5) [1], so the perpendicular bisector is y = 5 [1] (c) Midpoint of BD = (4, 5) and gradient BD = (3 − 7)/(6 − 2) = −1 [1]. Perpendicular gradient 1: y − 5 = x − 4, so y = x + 1 [1]. y = 5 meets x = 4 at (4, 5), and 4 + 1 = 5, so the bisector passes through (4, 5) [1] (d) The part of x = 4 from (4, 0) to (4, 5) [1] (e) y = x + 1 meets x = 0 at (0, 1) [1]. Area = rectangle 12 × 5 minus the triangle with corners (0, 1), (0, 5) and (4, 5): 60 − (1/2)(4)(4) [1] = 52 km² [1] Examiner insight: In (e) the corner (0, 1) must come from the edge equation, not a sketch, to earn the method mark.

Where marks are usually lost

• Using the segment’s gradient in the bisector equation instead of the perpendicular gradient.
• Giving y = mx + c when ax + by + d = 0 with integer coefficients is asked for.
• Stating a closest site with no distances or edge test to back it up.
• Averaging nearby values instead of taking the nearest site’s value in interpolation.
• Choosing the wrong vertex in a toxic waste problem, or not calculating the distance at every vertex.
• Rounding a vertex before finding a distance, which moves the answer in the third figure.

Next steps

• Revision notes for a fast recap of the methods
• Study guide for full explanations and worked examples
• IB DP Maths AI course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021 – syllabus sections SL 3.5 and SL 3.6.