IB Maths AI SL Topic 3 — Geometry Toolkit Paper 1 & 2 Coordinate basics ~7 min read

Coordinate Geometry

Given two points on a plane, three quantities describe the segment joining them: the midpoint (its centre), the distance (its length), and the gradient (its steepness). All three formulae are in the IB formula booklet — you don’t memorise them, but you must use them quickly and correctly. They underpin every later geometry topic: perpendicular bisectors, equations of lines, and trigonometry on the plane.

📘 What you need to know

Midpoint of A(x₁, y₁) and B(x₂, y₂) is the average of the two coordinates: M = ((x₁+x₂)/2, (y₁+y₂)/2).
Distance between the two points is the length of the segment: d = √((x₁−x₂)² + (y₁−y₂)²). This is Pythagoras’ theorem on the plane.
Gradient (slope) is rise over run: m = (y₂−y₁)/(x₂−x₁).
The line segment from A to B is written [AB]. Its length is written AB (no brackets).
Signs matter: positive gradient slopes up to the right; negative slopes down. Squaring in the distance formula removes signs, so subtraction order doesn’t matter for distance.
Order matters for gradient: but pick the SAME first point for both numerator and denominator — (y₂−y₁) and (x₂−x₁) must be in the same order.

The three formulae on one diagram

The picture below shows all three calculations for two points A(−3, −2) and B(5, 4). The horizontal leg is x₂−x₁ = 8, the vertical leg is y₂−y₁ = 6, and Pythagoras gives the length 10. The midpoint sits at the centre, and the gradient is rise/run = 6/8 = 3/4.

Two points A(−3, −2) and B(5, 4) on a coordinate plane. Orange dashed: the right-angled triangle giving the run (8) and rise (6). Green solid: the segment of length 10. The orange dot is the midpoint M(1, 1).

The three formulae — in the formula booklet midpoint = x₁+x₂2 , y₁+y₂2
d = √((x₁−x₂)² + (y₁−y₂)²)
m = y₂−y₁x₂−x₁

How to use them — common pitfalls

Midpoint: just average the xs and the ys separately. Negative coordinates feed in directly: (−3 + 5)/2 = 1.

Distance: squaring removes signs, so subtraction order doesn’t matter — (3 − 8)² = (−5)² = 25 = (8 − 3)². Gradient: order DOES matter, but only that you’re consistent — if B goes first in the numerator, it must go first in the denominator too. Otherwise the sign flips.

Quick sign check for gradient: positive when the line goes up to the right; negative when it goes down to the right. A rough sketch tells you the sign before you start.

Working backwards from the midpoint

Given the midpoint M and one endpoint, rearrange to find the other: x₂ = 2M_x − x₁ and y₂ = 2M_y − y₁ (twice the midpoint minus the known endpoint).

🧭 Recipe — any coordinate-geometry problem

Label your points: write down (x₁, y₁) and (x₂, y₂) above the coordinates. This avoids sign mistakes.
Identify which formula you need: midpoint (centre), distance (length), or gradient (slope).
Substitute the values carefully, especially for negative coordinates — use brackets: −(−3) = 3.
Simplify and evaluate. For distance, square first, then add, then take the root. For gradient, simplify the fraction.
State the answer with correct units or context: “distance = 10 units” or “the midpoint is (1, 1)”. A quick sketch confirms the answer makes sense.

Worked examples

WE 1

Midpoint — halfway between two cell towers

Two cell towers stand at P(2, 7) and Q(8, −3) on a survey grid (units in km). A new transmitter is to be installed at the midpoint of P and Q. Find its coordinates.

Step 1 — label the points (x₁, y₁) = (2, 7), (x₂, y₂) = (8, −3) Step 2 — average each coordinate Mₓ = (2 + 8)/2 = 10/2 = 5 Mᵢ = (7 + (−3))/2 = 4/2 = 2 M = (5, 2) notice how the negative coordinate just enters the sum naturally: 7 + (−3) = 4, then divide by 2. No special treatment.

WE 2

Distance — straight-line route between two parks

Park R is at coordinates (−1, 4) and Park S is at (11, 9), measured in km on a city grid. Find the straight-line distance between them.

Step 1 — label and find dx, dy dx = x₂ − x₁ = 11 − (−1) = 12 dy = y₂ − y₁ = 9 − 4 = 5 Step 2 — Pythagoras d = √(12² + 5²) = √(144 + 25) = √169 d = 13 km distance RS = 13 km (12, 5, 13) is a Pythagorean triple — like (3, 4, 5) and (5, 12, 13) it shows up often in exam questions because the answer comes out as a whole number.

WE 3

Gradient — slope of a ramp

A skateboard ramp has its two ends at points M(−2, 5) and N(6, 1). Find the gradient of the line through M and N, and state whether the ramp slopes up or down going from M to N.

Step 1 — label and substitute (x₁, y₁) = (−2, 5), (x₂, y₂) = (6, 1) m = (y₂ − y₁)/(x₂ − x₁) Step 2 — evaluate carefully with the negatives m = (1 − 5)/(6 − (−2)) = −4/8 = −1/2 m = −1/2 (down to the right) a negative gradient means y DECREASES as x increases. Going from M to N, the height drops from 5 to 1 — the ramp slopes downwards. Magnitude 1/2 means 1 unit down for every 2 units right.

WE 4

All three at once — hiking trail

A hiking trail runs in a straight line from camp A(−3, −2) to camp B(5, 4), with coordinates in km. Find (a) the midpoint of the trail, (b) the length of the trail, (c) the gradient of the trail.

(a) midpoint Mₓ = (−3 + 5)/2 = 1 Mᵢ = (−2 + 4)/2 = 1 M = (1, 1) (b) distance d = √((5−(−3))² + (4−(−2))²) = √(8² + 6²) = √(64+36) = √100 = 10 km (c) gradient m = (4 − (−2))/(5 − (−3)) = 6/8 = 3/4 (a) M(1, 1) · (b) 10 km · (c) m = 3/4 notice this is exactly the segment shown in the diagram at the top of this note. (6, 8, 10) is another Pythagorean triple — spot it and the distance is instant.

WE 5

Working backwards — find the missing endpoint

The midpoint of segment [PQ] is M(4, −1). The endpoint P has coordinates (7, 3). Find the coordinates of Q.

Step 1 — set up using the midpoint formula Mₓ = (Pₓ + Qₓ)/2 ⇒ (7 + Qₓ)/2 = 4 Mᵢ = (Pᵢ + Qᵢ)/2 ⇒ (3 + Qᵢ)/2 = −1 Step 2 — solve each equation 7 + Qₓ = 8 ⇒ Qₓ = 1 3 + Qᵢ = −2 ⇒ Qᵢ = −5 Step 3 — check by averaging midpoint of (7,3) and (1,−5) = (4, −1) ✓ Q = (1, −5) shortcut: Q = 2M − P. So Qₓ = 2(4) − 7 = 1, Qᵢ = 2(−1) − 3 = −5. Same answer, faster.

WE 6

Perimeter of a triangular plot

A triangular plot of land has vertices at A(1, 2), B(7, 10), C(13, 2), coordinates in metres. Find the perimeter of the plot.

Step 1 — length AB AB = √((7−1)² + (10−2)²) = √(36 + 64) = √100 = 10 Step 2 — length BC BC = √((13−7)² + (2−10)²) = √(36 + 64) = √100 = 10 Step 3 — length AC (horizontal) AC = √((13−1)² + 0²) = 12 Step 4 — perimeter P = 10 + 10 + 12 = 32 m perimeter = 32 m since AC is horizontal (both y = 2), its length is just |13 − 1| = 12. The triangle is isosceles: AB = BC. Spotting a horizontal/vertical side saves a Pythagoras calc.

💡 Top tips

Label (x₁, y₁) and (x₂, y₂) above each point: prevents sign errors when one coordinate is negative.
Recognise Pythagorean triples: (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17). If the legs match one, the distance is whole — no calculator needed.
Use brackets around negatives: write (−3) − (−5), not −3 − −5. Brackets stop the double-negative trap.
Distance is symmetric, gradient is not: AB = BA for length, but in gradient, putting B first in the numerator means putting B first in the denominator too.
Sketch the two points: a quick scribble shows which way the line slopes, so you can sanity-check the sign of your gradient and the rough size of your distance.

⚠ Common mistakes

Subtracting instead of adding in the midpoint: midpoint uses (x₁ + x₂)/2, NOT (x₁ − x₂)/2. Average means add then divide.
Forgetting to square before adding in the distance formula: d ≠ |x₂−x₁| + |y₂−y₁|. You must square, add, then square-root.
Sign errors with negative coordinates: 5 − (−3) = 8, not 2. Two minus signs make a plus.
Flipping numerator and denominator in gradient: m is rise OVER run (vertical change over horizontal change), not run over rise. Flipping gives the reciprocal — usually a perpendicular line, not the one you want.
Mixing the order of points in the gradient formula: (y₂−y₁) on top requires (x₂−x₁) on the bottom — same order. Swap one and you flip the sign.

Up next: Perpendicular Bisectors. Combining all three skills from this note — midpoint, gradient, and the equation of a straight line — you’ll learn how to find the line that cuts a segment exactly in half AT a right angle. It’s the foundation for Voronoi diagrams later in the course.

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