IB Maths AI HL Voronoi Diagrams Paper 1 & 2 Perp bisectors & cells ~8 min read

Drawing Voronoi Diagrams

A Voronoi diagram divides a plane into cells, one per site (a fixed point), where each cell contains every location closer to that site than to any other. The edges are perpendicular bisectors between pairs of sites; the vertices are points equidistant from three sites. Most exam questions ask you to find the equation of an edge or locate a missing site — both reduce to standard perpendicular-bisector work.

📘 What you need to know

Site, cell, edge, vertex: a site is a fixed point; its cell is the region closer to it than to any other site; edges are perpendicular bisectors between two sites; vertices are intersections of three perpendicular bisectors.
Each edge is the perpendicular bisector between the two sites in the neighbouring cells — the boundary is the set of points equidistant from both.
Each vertex is equidistant from three sites — this is what allows you to find vertex coordinates by intersecting two perpendicular bisectors.
Midpoint formula: (x₁+x₂2, y₁+y₂2).
Perpendicular gradient: if the line through two sites has gradient y₂−y₁x₂−x₁, the perpendicular bisector has gradient −x₂−x₁y₂−y₁.
Finding a missing site: pick an edge of the missing cell, reflect a neighbouring site across that edge; the reflected point is the missing site.

Cells, edges, and the perpendicular bisector

Every edge in a Voronoi diagram is the perpendicular bisector of the segment joining the two sites in the neighbouring cells. Every point on that edge is equidistant from those two sites, which is exactly the boundary condition: cross the edge and you switch which site is nearest. Three edges meet at a vertex, which is then equidistant from three sites simultaneously — the circumcentre of the triangle formed by those three sites.

Three sites A(2, 6), B(8, 6), C(5, 1) define a Voronoi diagram with three cells meeting at the vertex V. Each edge (in teal) is a perpendicular bisector; the vertex is equidistant from all three sites.

Edge between sites (x₁, y₁) and (x₂, y₂) pass through M = (x₁+x₂2, y₁+y₂2) with gradient m_⊥ = −x₂−x₁y₂−y₁ use y − y_M = m_⊥(x − x_M), then rearrange to ax + by + d = 0 (integer coefficients)

Finding a vertex or a missing site

A vertex is the intersection of any two perpendicular bisectors (the third will pass through it automatically because it’s equidistant from all three sites). To find a vertex: write the equations of two edges and solve them simultaneously. To find a missing site: pick a known site in a neighbouring cell, and reflect it across the shared edge. The reflection sits on the perpendicular through the original site, the same distance on the other side — that’s the missing site.

Reflection shortcut: to reflect point P across a line, drop a perpendicular from P to the line, find the foot F, then the reflection P′ satisfies F = midpoint of PP′, so P′ = 2F − P.

🧭 Recipe — equation of an edge

Identify the two sites whose cells share the edge.
Compute the midpoint of the segment between them.
Compute the gradient of the segment.
Perpendicular gradient is the negative reciprocal.
Use point-slope form with the midpoint and perpendicular gradient, then rearrange to ax + by + d = 0 with integer a, b, d.

Worked examples

WE 1

Equation of an edge between two sites

Two sites in a Voronoi diagram are P(0, 0) and Q(8, 6). Find the equation of the edge between the two cells, giving your answer in the form ax + by + d = 0 with integer a, b, d.

midpoint of PQ M = ((0+8)/2, (0+6)/2) = (4, 3) gradient of PQ m = (6 − 0)/(8 − 0) = 3/4 perpendicular gradient m⊥ = −4/3 point-slope through M y − 3 = −4/3 (x − 4) 3y − 9 = −4x + 16 4x + 3y − 25 = 0 check: M = (4, 3) gives 4·4 + 3·3 − 25 = 16 + 9 − 25 = 0 ✓

WE 2

Edge with negative coordinates

Two sites are M(−2, 3) and N(4, −1). Find the equation of the perpendicular bisector that forms the edge between their cells, in the form ax + by + d = 0 with integer coefficients.

midpoint of MN ((−2+4)/2, (3+(−1))/2) = (1, 1) gradient of MN m = (−1 − 3)/(4 − (−2)) = −4/6 = −2/3 perpendicular gradient m⊥ = 3/2 point-slope through (1, 1) y − 1 = 3/2 (x − 1) 2y − 2 = 3x − 3 3x − 2y − 1 = 0 check at (1,1): 3 − 2 − 1 = 0 ✓

WE 3

Vertex of a Voronoi diagram (three sites forming a right triangle)

A Voronoi diagram has three sites P(0, 0), Q(10, 0), and R(0, 10). Find the coordinates of the vertex of the diagram — the point equidistant from all three sites.

perpendicular bisector of PQ (horizontal segment) midpoint (5, 0), PQ horizontal → ⊥ is vertical ⊥PQ: x = 5 perpendicular bisector of PR (vertical segment) midpoint (0, 5), PR vertical → ⊥ is horizontal ⊥PR: y = 5 vertex = intersection V = (5, 5) since △PQR is right-angled at P, the vertex is the midpoint of the hypotenuse QR.

WE 4

Vertex from two edge equations

On a Voronoi diagram, the edge between cells P and Q has equation 2x + y = 10, and the edge between cells Q and R has equation x − y = 2. Find the coordinates of the vertex where these two edges meet.

solve the simultaneous equations 2x + y = 10 x − y = 2 add the two 3x = 12 → x = 4 y = x − 2 = 2 V = (4, 2) two perpendicular bisectors are enough — the third will pass through V automatically.

WE 5

Locate a missing site by reflection

A Voronoi diagram has site P(3, 1). The edge of P’s cell that separates P from the missing site Q has equation y = −x + 8. Find the coordinates of Q.

edge has gradient −1, so PQ has gradient +1 line through P(3, 1) with gradient 1: y = x − 2 foot of perpendicular F = intersection of edge and PQ x − 2 = −x + 8 → 2x = 10 → x = 5 y = 3, so F = (5, 3) Q = 2·F − P (reflect P across the edge) Q = (2·5 − 3, 2·3 − 1) = (7, 5) Q = (7, 5) check: midpoint of PQ = (5, 3) lies on the edge, and PQ ⊥ edge ✓

WE 6

Vertex of three sites with a vertical symmetry axis

A Voronoi diagram has sites P(2, 2), Q(8, 2), and R(5, 8). Find the coordinates of the vertex equidistant from all three sites.

⊥PQ: PQ is horizontal, midpoint (5, 2) ⊥PQ: x = 5 ⊥PR: midpoint (3.5, 5), gradient PR = 2 m⊥ = −1/2 y − 5 = −1/2 (x − 3.5) substitute x = 5 into ⊥PR y − 5 = −1/2 (5 − 3.5) = −0.75 y = 4.25 = 17/4 V = (5, 17/4) check distances: PV² = 9 + (9/4)² = 9 + 81/16 = 225/16; QV² same by symmetry; RV² = 0 + (8 − 17/4)² = (15/4)² = 225/16 ✓

💡 Top tips

Always start with midpoint and gradient — these are the two building blocks for every edge equation.
Horizontal/vertical sites simplify things: if the segment is horizontal, the perpendicular bisector is vertical (and vice versa) — no need for gradients.
Two bisectors are enough for a vertex: the third will pass through it because it’s equidistant from all three sites.
For a missing site, pick the simplest edge — a horizontal or vertical edge gives the cleanest reflection.
Clear fractions early: when writing the edge in ax + by + d = 0 form, multiply through by the denominators to get integer coefficients.

⚠ Common mistakes

Using the segment gradient instead of the perpendicular gradient — the edge is perpendicular to the segment, so take the negative reciprocal.
Sign errors with negative coordinates in midpoint or gradient calculations — write each term explicitly.
Forgetting to clear fractions — the question often demands integer a, b, d.
Reflecting a site across the wrong edge — for a missing site, pick the edge that borders its cell, not some unrelated edge.
Mixing up which sites share an edge — only the two sites in the neighbouring cells matter for that edge.

Next up — Interpreting Voronoi Diagrams. With the construction in hand, the focus shifts to using a Voronoi diagram to answer practical questions: which site is closest to a given point, what’s the shortest distance, and which site’s data should be used to predict outcomes at a new location (nearest-neighbour interpolation).

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