IB Maths AI HL Geometry Toolkit Paper 1 & 2 5-step procedure ~9 min read

Perpendicular Bisectors

The perpendicular bisector of a line segment cuts it in half at a right angle — and it has a powerful geometric meaning: every point on it is equidistant from the two endpoints. Finding its equation combines all three coordinate-geometry formulas into a clean 5-step procedure.

📘 What you need to know

Definition: the perpendicular bisector of [AB] is the line that passes through the midpoint of [AB] at a right angle to [AB].
Equidistance property: every point on the perpendicular bisector of [AB] is the same distance from A as it is from B.
Perpendicular gradient rule: if a line has gradient m, then a perpendicular line has gradient m_⊥ = −1/m.
5-step procedure: midpoint → gradient of AB → perpendicular gradient → equation through midpoint → rearrange.
Special cases: if AB is horizontal, the perp bisector is vertical (x = midpoint’s x); if AB is vertical, the perp bisector is horizontal.
Forms of the answer: y = mx + c or ax + by + d = 0. Both are in the formula booklet; the exam may specify which.

The 5-step procedure

To find the equation of the perpendicular bisector of [AB]: first compute the midpoint — the perp bisector must pass through it. Next compute the gradient of AB. Flip it and change sign to get the perpendicular gradient. Use the point-slope form y − y₁ = m(x − x₁) with the midpoint coordinates and the perpendicular gradient. Finally rearrange to whatever form the question asks for. The whole procedure is mechanical — the only place errors creep in is arithmetic with negatives, especially when flipping the gradient.

The perpendicular bisector of [AB] passes through the midpoint at a right angle. Every point on it (like P) is equidistant from A and B.

Perpendicular gradient rule m_⊥ = −1/m · point-slope form: y − y₁ = m(x − x₁) use the midpoint as (x₁, y₁) and the perpendicular gradient as m

The equidistance property

The geometric heart of the perpendicular bisector is that it’s exactly the locus (set of points) equidistant from two given points. This is what makes perpendicular bisectors show up in applied problems: drawing a boundary between two cell towers where signal strengths are equal, finding the location for a fire station equidistant from two towns, or locating the circumcentre of a triangle (the point equidistant from all three vertices, where any two perpendicular bisectors intersect). To check whether a point lies on the bisector, you can either substitute into the bisector’s equation or just compute the two distances and compare.

Equidistance test: to find the perpendicular bisector without using gradients, set up the equation |PA|² = |PB|² for a general point P(x, y), then expand and simplify. The squared distances ensure you keep the equation polynomial and the squares of x and y cancel, leaving a linear equation — exactly the perpendicular bisector.

🧭 Recipe — perpendicular bisector of [AB]

Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2).
Gradient of AB: m = (y₂−y₁)/(x₂−x₁).
Perpendicular gradient: m_⊥ = −1/m (flip and change sign).
Equation through midpoint: y − M_y = m_⊥(x − M_x).
Rearrange to y = mx + c or ax + by + d = 0 as requested.

Worked examples

WE 1

Standard perpendicular bisector in y = mx + c form

Points P(2, 1) and Q(8, 5). Find the equation of the perpendicular bisector of [PQ] in the form y = mx + c.

Step 1: midpoint M = ((2+8)/2, (1+5)/2) = (5, 3) Step 2: gradient PQ m = (5−1)/(8−2) = 4/6 = 2/3 Step 3: perpendicular gradient m_⊥ = −1/(2/3) = −3/2 Step 4: equation through M y − 3 = −3/2 (x − 5) y = −3x/2 + 15/2 + 3 y = −3x/2 + 21/2

WE 2

Form ax + by + d = 0 with integer coefficients

Points A(−3, 2) and B(5, 8). Find the perpendicular bisector of [AB] in the form ax + by + d = 0 with integer coefficients.

Step 1: midpoint M = ((−3+5)/2, (2+8)/2) = (1, 5) Step 2: gradient AB m = (8−2)/(5−(−3)) = 6/8 = 3/4 Step 3: perpendicular gradient m_⊥ = −1/(3/4) = −4/3 Step 4: equation through M y − 5 = −4/3 (x − 1) Step 5: multiply both sides by 3 3y − 15 = −4(x − 1) = −4x + 4 4x + 3y − 19 = 0 4x + 3y − 19 = 0 check at M(1, 5): 4(1) + 3(5) − 19 = 4 + 15 − 19 = 0 ✓

WE 3

Verify the equidistance property

Given A(1, 4) and B(7, 2): (a) Find the perpendicular bisector of [AB]. (b) Verify that P(5, 6) lies on this bisector. (c) Show that P is equidistant from A and B.

(a) midpoint M = (4, 3) gradient AB = (2−4)/(7−1) = −1/3 perp gradient = 3 y − 3 = 3(x − 4) y = 3x − 9 (b) substitute P(5, 6) 3(5) − 9 = 15 − 9 = 6 ✓ P lies on the bisector (c) distances PA = √((5−1)² + (6−4)²) = √(16+4) = √20 PB = √((5−7)² + (6−2)²) = √(4+16) = √20 PA = PB = 2√5 ✓

WE 4

Circumcentre via two perpendicular bisectors

A triangle has vertices A(0, 0), B(6, 0), and C(0, 8). The circumcentre is the point equidistant from all three vertices. Find it by intersecting the perpendicular bisectors of two sides.

perp bisector of [AB] midpoint (3, 0); AB horizontal → perp is vertical x = 3 perp bisector of [AC] midpoint (0, 4); AC vertical → perp is horizontal y = 4 intersect circumcentre = (3, 4) verify equidistance to A: √(9+16) = √25 = 5 to B: √(9+16) = 5 to C: √(9+16) = 5 all distances = 5 ✓ (3, 4, 5) Pythagorean triple shows up nicely.

WE 5

Applied: cell-tower equidistance boundary

Two cell towers stand at A(−2, 3) and B(8, 1) (coordinates in km). Engineers want the boundary between cells — the locus of points equidistant from both towers. Find its equation in the form ax + by + d = 0.

the boundary IS the perpendicular bisector of [AB] Step 1: midpoint M = ((−2+8)/2, (3+1)/2) = (3, 2) Step 2: gradient AB m = (1−3)/(8−(−2)) = −2/10 = −1/5 Step 3: perp gradient m_⊥ = 5 Step 4: equation through M y − 2 = 5(x − 3) y = 5x − 13 5x − y − 13 = 0

WE 6

Reverse: find an unknown coordinate

Given A(2, 4) and B(6, 10). A point P(k, 9) lies on the perpendicular bisector of [AB]. Find k.

Step 1: midpoint M = (4, 7) Step 2: gradient AB m = (10−4)/(6−2) = 6/4 = 3/2 Step 3: perp gradient m_⊥ = −2/3 Step 4: equation through M y − 7 = −2/3 (x − 4) substitute P(k, 9) 9 − 7 = −2/3 (k − 4) 2 = −2/3 (k − 4) ⇒ k − 4 = −3 k = 1 verify by distances PA = √(1 + 25) = √26 PB = √(25 + 1) = √26 ✓

💡 Top tips

Always start with the midpoint — the perp bisector passes through it, so you need the coordinates for Step 4.
Flip and change sign for the perpendicular gradient: 2/3 → −3/2; −4 → 1/4; 5 → −1/5.
Multiply through to clear fractions when the question wants integer-coefficient form ax + by + d = 0.
Check by substituting the midpoint into your final equation — it must satisfy it (gives 0 = 0).
Special cases: horizontal AB → vertical perp bisector (x = constant); vertical AB → horizontal perp bisector.

⚠ Common mistakes

Using gradient of AB instead of its negative reciprocal — that gives a line parallel to AB, not perpendicular.
Forgetting to use the midpoint in Step 4 and using an endpoint instead — the resulting line won’t bisect the segment.
Sign error when flipping the gradient: −1/(m) means flip the fraction AND change the sign.
Not rearranging to the requested form — check the question for whether y = mx + c or ax + by + d = 0 is wanted.
Missing the special case: if AB is horizontal, the perp gradient isn’t −1/0 — the perp bisector is vertical (x = midpoint’s x-coordinate).

Next up: Arcs & Sectors Using Degrees — finding the length of an arc and area of a sector using fractions of a full circle.

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Perpendicular Bisectors

📘 What you need to know

The 5-step procedure

The equidistance property

🧭 Recipe — perpendicular bisector of [AB]

Worked examples

💡 Top tips

⚠ Common mistakes

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