IB Maths AA SL Topic 3 — Geometry & Trig Paper 1 & 2 ~7 min read

Coordinate Geometry

Three formulas. That’s it. Given two points on the plane, you’ll always need to find one of these: the midpoint, the distance, or the gradient. Get fluent at substituting, and this topic gives you free marks every paper.

📘 What you need to know

Midpoint of two points = average of their x‘s and average of their y‘s.
Distance between two points = Pythagoras’ theorem in disguise.
Gradient of a line through two points = “rise over run” = change in y / change in x.
Midpoint and distance are in the formula booklet under prior learning; gradient is under functions.
Always label your points (x1, y1) and (x2, y2) before substituting — the #1 way to avoid sign errors.

The three formulas at a glance

Memorise these. They’ll come up in every Geometry paper at SL.

📍

Midpoint

(x1 + x22, y1 + y22)

✓ in formula booklet

📏

Distance

d = √((x1 − x2)² + (y1 − y2)²)

✓ in formula booklet

📈

Gradient

m = y2 − y1x2 − x1

✓ in formula booklet

All three formulas are given in the IB formula booklet, but you should know them by heart. You shouldn’t have to flip pages mid-paper — that wastes time on easy marks.

Two points · Three things to find

x = run Δy = rise d A(x₁, y₁) B(x₂, y₂) M (midpoint)

1. The midpoint formula

The midpoint of two points is just the average of each coordinate. Add the two x‘s and divide by 2. Add the two y‘s and divide by 2. Done.

Midpoint of (x1, y1) and (x2, y2) M = (x1 + x22, y1 + y22) ✓ in formula booklet (prior learning)

Quick example: Midpoint of (4, −2) and (6, 8) is (4 + 62, −2 + 82) = (5, 3).

2. The distance formula

The distance between two points is the length of the line segment joining them. The formula is just Pythagoras’ theorem applied on the coordinate grid.

Distance between (x1, y1) and (x2, y2) d = √( (x1 − x2)² + (y1 − y2)² ) ✓ in formula booklet (prior learning)

🤔 Why is this just Pythagoras?

Drop a horizontal line and a vertical line from your two points. You’ve made a right-angled triangle whose horizontal side is (x1 − x2) and vertical side is (y1 − y2). The distance is the hypotenuse, so d² = (Δx)² + (Δy)². Take the square root and you’ve got the formula.

Order doesn’t matter: (x1 − x2)² = (x2 − x1)² because squaring removes the sign. So you can subtract either way round.

In IB notation, the line segment between points A and B is written [AB] (with square brackets) and its length is just AB. Use the right notation in your written answers.

3. The gradient formula

The gradient (or slope) of a line tells you how steep it is. It’s rise over run — the change in y divided by the change in x.

Gradient of line through (x1, y1) and (x2, y2) m = y2 − y1x2 − x1 ✓ in formula booklet (functions)

Order matters here — but only in the sense that you must keep things consistent. If you put y2 on top, you must put x2 on the bottom. Don’t mix them up.

What does the sign of m tell you?

Positive

m > 0 · slopes up

Negative

m < 0 · slopes down

Zero

m = 0 · horizontal

Undefined

vertical line

A vertical line has no gradient — the formula gives division by zero. Watch for this in problems involving lines like x = 4.

Worked examples

WE 1

Find the midpoint

Find the midpoint of P(7, −3) and Q(−1, 5).

Step 1: Label the coordinates P: x1 = 7, y1 = −3 Q: x2 = −1, y2 = 5 Step 2: Average each coordinate x: 7 + (−1)2 = 62 = 3 y: −3 + 52 = 22 = 1 Midpoint = (3, 1)

WE 2

Find the distance between two points

Find the distance between C(2, 1) and D(8, 9).

Step 1: Label and substitute d = √( (2 − 8)² + (1 − 9)² ) Step 2: Compute = √( (−6)² + (−8)² ) = √(36 + 64) = √100 d = 10 units classic 3-4-5 triangle scaled by 2

WE 3

Find the gradient

Find the gradient of the line through E(−2, 3) and F(4, −9).

Step 1: Label and substitute E: x1 = −2, y1 = 3 F: x2 = 4, y2 = −9 Step 2: Apply the formula m = −9 − 34 − (−2) = −126 m = −2 negative → line slopes down

WE 4

All three formulas in one question (SME-style)

Point A has coordinates (3, −4) and point B has coordinates (−5, 2). Find:

(i) the distance AB (ii) the gradient of AB (iii) the midpoint of [AB]

(i) Distance d = √( (3 − (−5))² + (−4 − 2)² ) = √(8² + (−6)²) = √(64 + 36) = √100 AB = 10 units (ii) Gradient m = 2 − (−4)−5 − 3 = 6−8 m = −34 (iii) Midpoint M = (3 + (−5)2, −4 + 22) = (−22, −22) M = (−1, −1)

WE 5

Reverse: find the missing endpoint

The midpoint of [PQ] is M(2, −1). Given that P(−3, 4), find the coordinates of Q.

Step 1: Set up the midpoint equations x-coord: −3 + x2 = 2 y-coord: 4 + y2 = −1 Step 2: Solve each −3 + x = 4 → x = 7 4 + y = −2 → y = −6 Q = (7, −6) just multiply both sides by 2 — never overthink it

💡 Top tips

Label your coordinates first. Write x1, y1, x2, y2 next to each pair before substituting — it eliminates 90% of sign errors.
Sketch a quick diagram. Even a rough plot of the two points reveals whether your gradient sign and rough distance look right.
Treat the distance formula as Pythagoras. If you ever forget it, sketch the triangle and compute Δx² + Δy², then take the square root.
Negative numbers — use brackets. Write 3 − (−5) instead of 3 − −5 to avoid confusion.
Leave answers exact when possible. If √50 comes up, leave it as 5√2 unless a decimal is asked for.
Use IB notation correctly. [AB] = the line segment, AB = the length, (AB) = the line through A and B.
Distance is always positive — if you get a negative number, you’ve made a sign error somewhere.

⚠ Common mistakes

Mixing up the order in the gradient formula. If you put y2 on top, you must put x2 on the bottom. Mismatching them gives the wrong sign.
Subtracting twice instead of squaring. Don’t forget the square in the distance formula — (x1 − x2) by itself isn’t enough.
Forgetting the square root after computing Δx² + Δy². The result before the root is just d², not d.
Sign errors with negative coordinates. Subtracting a negative becomes adding — −4 − (−6) = −4 + 6 = 2. Always use brackets.
Adding instead of averaging the midpoint. The midpoint is the average, so don’t forget to divide by 2.
Saying a vertical line has gradient 0. Vertical lines have an undefined gradient — horizontal lines have gradient 0.
Confusing midpoint with gradient when both come up. Read the question carefully — they look superficially similar but use different formulas.

These three formulas come up in every Geometry paper. Get fluent with them now and you’ll save brainpower for the harder questions later in the paper.

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Coordinate Geometry

📘 What you need to know

The three formulas at a glance

1. The midpoint formula

2. The distance formula

🤔 Why is this just Pythagoras?

3. The gradient formula

What does the sign of m tell you?

Worked examples

💡 Top tips

⚠ Common mistakes

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