IB Maths AI SL Topic 3 — Geometry Toolkit Paper 1 & 2 Equidistant lines ~6 min read

Perpendicular Bisectors

The perpendicular bisector of a segment [AB] is the line that (i) passes through the midpoint of A and B, and (ii) meets [AB] at 90°. Every point on it is equidistant from A and B. That property makes it the foundation for Voronoi diagrams later in the course — “the boundary between two territories”.

📘 What you need to know

Two conditions: perpendicular to [AB] AND passes through the midpoint of A and B.
Perpendicular gradient rule: if m_AB = m, then the perpendicular gradient is m_⊥ = −1/m. Their product is −1.
Five-step method: midpoint → gradient of [AB] → negative reciprocal → point-gradient line → rearrange to required form.
Vertical segment (x₁ = x₂) gives a horizontal bisector y = M_y. Horizontal segment (y₁ = y₂) gives a vertical bisector x = M_x.
Equidistance property: any point P on the perpendicular bisector satisfies PA = PB.
Forms accepted: y = mx + c OR ax + by + d = 0. Read the question for which form is required.

How to find one — in one picture

Green: the segment [AB]. Orange: its perpendicular bisector, crossing the midpoint M(5, 3) at a right angle. Any point P on the orange line is the same distance from A and from B — that’s the defining property.

Perpendicular gradient rule m_AB · m_⊥ = −1 ⇒ m_⊥ = − 1m_AB

line through M with gradient m_⊥: y − M_y = m_⊥ (x − M_x)

🧭 Recipe — perpendicular bisector of [AB]

Midpoint of A and B: M = ((x₁+x₂)/2, (y₁+y₂)/2).
Gradient of [AB]: m = (y₂−y₁)/(x₂−x₁).
Perpendicular gradient: m_⊥ = −1/m. Flip and negate.
Substitute into y − M_y = m_⊥(x − M_x).
Rearrange to the required form — y = mx + c or ax + by + d = 0.

Special cases: if [AB] is vertical, the bisector is y = M_y. If [AB] is horizontal, the bisector is x = M_x. No gradient calculation needed.

Worked examples

WE 1

Standard case — positive gradient

Find the equation of the perpendicular bisector of [AB], where A(2, 1) and B(8, 5). Give your answer in the form ax + by + d = 0.

Step 1 — midpoint M = ((2+8)/2, (1+5)/2) = (5, 3) Step 2 — gradient of AB m = (5−1)/(8−2) = 4/6 = 2/3 Step 3 — perpendicular gradient m⊥ = −1/(2/3) = −3/2 Step 4 — line through M with gradient m⊥ y − 3 = −(3/2)(x − 5) Step 5 — rearrange to ax + by + d = 0 2(y − 3) = −3(x − 5) 2y − 6 = −3x + 15 3x + 2y − 21 = 0 3x + 2y − 21 = 0 check: midpoint (5,3) → 3(5)+2(3)−21 = 15+6−21 = 0 ✓

WE 2

Vertical segment — horizontal bisector

Find the perpendicular bisector of [PQ] where P(3, 2) and Q(3, 8).

Spot the special case x₁ = x₂ = 3 → [PQ] is VERTICAL Midpoint M = (3, (2+8)/2) = (3, 5) Bisector is horizontal through M y = 5 y = 5 no need to fiddle with “gradient undefined” — just recognise: vertical segment → horizontal bisector with y equal to midpoint y-coordinate.

WE 3

Horizontal segment — vertical bisector

Find the perpendicular bisector of [AB] where A(−4, 7) and B(6, 7).

Spot the special case y₁ = y₂ = 7 → [AB] is HORIZONTAL Midpoint M = ((−4+6)/2, 7) = (1, 7) Bisector is vertical through M x = 1 x = 1 horizontal segment → vertical bisector. The bisector equation only mentions x; it has no y in it.

WE 4

Fire station location — equidistant from two towns

A fire station is to be built equidistant from town X(2, −3) and town Y(10, 5). Find the equation of the line along which the station could be located, in the form y = mx + c.

Step 1 — midpoint M = ((2+10)/2, (−3+5)/2) = (6, 1) Step 2 — gradient XY m = (5−(−3))/(10−2) = 8/8 = 1 Step 3 — perpendicular gradient m⊥ = −1/1 = −1 Step 4 — line through M y − 1 = −1(x − 6) y = −x + 6 + 1 y = −x + 7 y = −x + 7 “equidistant from X and Y” is the defining feature of the perpendicular bisector. Any fire station on this line is the same distance from both towns.

WE 5

Fractional gradient — phone tower between villages

A mobile-phone tower is to serve village A(−2, 4) and village B(6, 8). Find the equation of the perpendicular bisector of [AB] in the form ax + by + d = 0.

Step 1 — midpoint M = ((−2+6)/2, (4+8)/2) = (2, 6) Step 2 — gradient AB m = (8−4)/(6−(−2)) = 4/8 = 1/2 Step 3 — perpendicular gradient m⊥ = −1/(1/2) = −2 Step 4 + 5 — line and rearrange y − 6 = −2(x − 2) y − 6 = −2x + 4 2x + y − 10 = 0 2x + y − 10 = 0 “flip and negate”: 1/2 → 2 → −2. The two gradients (1/2 and −2) multiply to −1 ✓

WE 6

Find an equidistant point on the x-axis

Find the point on the x-axis that is equidistant from A(1, 3) and B(9, 7).

Step 1 — set up: equidistant ⇒ on the perp bisector M = ((1+9)/2, (3+7)/2) = (5, 5) m_AB = (7−3)/(9−1) = 4/8 = 1/2 m⊥ = −2 Step 2 — perp bisector equation y − 5 = −2(x − 5) y = −2x + 15 Step 3 — meet the x-axis (y = 0) 0 = −2x + 15 x = 15/2 = 7.5 (7.5, 0) the x-axis is the line y = 0. The equidistant point is where the perpendicular bisector crosses it. Same trick works for finding a point on the y-axis (set x = 0).

💡 Top tips

“Flip and negate”: m = 3/4 → m⊥ = −4/3. m = −2 → m⊥ = 1/2. m = 1 → m⊥ = −1.
Spot vertical/horizontal segments first: skips all the gradient work. x₁ = x₂ → horizontal bisector; y₁ = y₂ → vertical bisector.
Use the midpoint as your “point” in y − y₀ = m(x − x₀), not A or B. The bisector passes through M, not the endpoints.
Always check: midpoint satisfies the equation. Substitute M_x, M_y into your final answer; you should get 0 (or both sides equal).
“Equidistant” is your trigger word — when it appears, perpendicular bisector is the tool.

⚠ Common mistakes

Using the gradient of [AB] for the bisector: the bisector has the perpendicular gradient. Don’t forget to flip and negate.
Just negating without flipping: m = 2/3 ↛ −2/3. Correct: m⊥ = −3/2. Both operations.
Substituting A or B instead of M in the line equation. The bisector goes through the MIDPOINT, not the endpoints.
Forgetting the special cases: a vertical segment has no defined gradient — don’t try −1/0. Just recognise: horizontal bisector y = Mᵢ.
Answer in the wrong form: the question says “ax + by + d = 0”, you give y = mx + c. Match the requested form.

Up next: Arcs & Sectors. Two formulae handle every circle question on AI SL: arc length = (θ/360)×2πr and sector area = (θ/360)×πr². Both are in the formula booklet — you only need to know how to apply them to slices, perimeters and shaded regions.

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