IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~6 min read
Coordinate Geometry
Coordinate geometry is the toolkit that links algebra and the plane. Three formulas do almost all the work: the midpoint of a line segment, the distance between two points, and the gradient of the line joining them. They appear constantly throughout the IB course — in trigonometry, vectors, calculus, and modelling — so getting the basics fast and accurate pays dividends everywhere.
📘 What you need to know
- Midpoint: the middle point of the segment joining (x1, y1) and (x2, y2) is the average of the coordinates.
- Distance: the length of [AB] comes from Pythagoras applied to the differences x2 − x1 and y2 − y1.
- Gradient: rise over run — the change in y divided by the change in x.
- Notation: [AB] denotes the line segment between A and B.
- All three formulas are in the formula booklet: midpoint and distance under “prior learning”; gradient under “functions”.
- Order doesn’t matter for distance: differences are squared, so the sign cancels. Order does matter for gradient — keep both subscripts consistent.
- Perpendicular lines: gradients multiply to −1. Parallel lines: gradients are equal.
- Use diagrams: a quick sketch catches sign errors and tells you which formula to reach for.
Midpoint of two points
The midpoint sits exactly halfway along the segment joining two points — equidistant from both ends. Each coordinate is just the average of the two coordinates of the same name.
Midpoint formula
M = ( x1 + x22 , y1 + y22 )
That’s it — average the x‘s, average the y‘s. The order of the points makes no difference, since addition is commutative.
Reverse use: if you know the midpoint and one endpoint, you can solve for the other endpoint by rearranging. For midpoint M(mx, my) and known endpoint P, the other endpoint is Q = (2mx − Px, 2my − Py).
Distance between two points
The distance is the length of the line segment joining the points. Drop a horizontal and vertical from each point and you get a right-angled triangle whose hypotenuse is the segment — Pythagoras finishes the job.
Distance formula
d = √( (x1 − x2)2 + (y1 − y2)2 )
The differences are squared, so it doesn’t matter which point you call (x1, y1) — negative differences become positive squares. Always take the positive square root: distance is non-negative.
If you forget the formula, redraw the triangle: horizontal leg = |Δx|, vertical leg = |Δy|, hypotenuse = distance. The formula is just Pythagoras.
Gradient between two points
The gradient measures the steepness — how much the y-value rises (or falls) per unit increase in x.
Gradient formula
m = y2 − y1x2 − x1
Order matters here in one specific way: the first y on the numerator must come from the same point as the first x on the denominator. You can swap both rows simultaneously and get the same answer, but mixing them up flips the sign.
Parallel lines
m1 = m2
same gradient, different intercepts
Perpendicular lines
m1 · m2 = −1
gradients are negative reciprocals
🧭 Recipe — coordinate geometry workflow
- Label the points clearly: write (x1, y1) above the first point and (x2, y2) above the second.
- Pick the right formula: midpoint for “halfway”, distance for “length” or “how far”, gradient for “slope” or “steepness”.
- Substitute carefully: keep negatives in brackets to avoid sign slips — (x2) − (x1), not x2 − x1 with signs floating.
- Simplify: leave fractions in lowest terms; leave surds exact unless asked for a decimal.
- Sanity check with a sketch: does the gradient sign match the line’s direction? Does the midpoint sit visually between the two endpoints?
Worked examples
WE 1Distance between two points
Find the distance between P(1, 7) and Q(4, 11).
Step 1: Label the coordinates
(x₁, y₁) = (1, 7); (x₂, y₂) = (4, 11)
Step 2: Substitute into d = √((x₁−x₂)² + (y₁−y₂)²)
d = √((1−4)² + (7−11)²)
d = √((−3)² + (−4)²) = √(9 + 16)
d = √25 = 5
d = 5 units
a 3-4-5 triangle hidden in the differences — clean exact answer
WE 2Gradient of a line through two points
Find the gradient of the line passing through M(−2, 8) and N(6, −4).
Step 1: Label coordinates
(x₁, y₁) = (−2, 8); (x₂, y₂) = (6, −4)
Step 2: Substitute into m = (y₂−y₁)/(x₂−x₁)
m = (−4 − 8)/(6 − (−2))
m = −12/8
simplify: divide top and bottom by 4
m = −3/2
negative gradient — line slopes downward from left to right, which matches: y went from 8 down to −4 as x increased
WE 3Midpoint of a line segment
Find the midpoint of the line segment joining C(7, −3) and D(−1, 9).
Step 1: Apply midpoint formula
M = ((x₁+x₂)/2, (y₁+y₂)/2)
Step 2: Substitute
M = ((7 + (−1))/2, (−3 + 9)/2)
M = (6/2, 6/2)
M = (3, 3)
a quick sketch confirms (3, 3) sits visually halfway between (7, −3) and (−1, 9)
WE 4Distance, gradient, and midpoint together
For the points E(2, 5) and F(8, −3), find (a) the length of [EF], (b) the gradient of the line EF, and (c) the midpoint of [EF].
(a) Distance
d = √((2−8)² + (5−(−3))²)
d = √((−6)² + 8²) = √(36 + 64)
d = √100 = 10
|EF| = 10 units
(b) Gradient
m = (−3 − 5)/(8 − 2) = −8/6
m = −4/3
(c) Midpoint
M = ((2+8)/2, (5+(−3))/2) = (10/2, 2/2)
M = (5, 1)
another clean Pythagorean triple — 6, 8, 10 — hiding behind the coordinates
WE 5Find an unknown endpoint from the midpoint
The midpoint of [PQ] is M(2, 4) and one endpoint is P(−3, 1). Find the coordinates of the other endpoint Q.
Step 1: Set up the midpoint equations
((−3 + Qx)/2, (1 + Qy)/2) = (2, 4)
Step 2: Solve for Qx
(−3 + Qx)/2 = 2 → −3 + Qx = 4 → Qx = 7
Step 3: Solve for Qy
(1 + Qy)/2 = 4 → 1 + Qy = 8 → Qy = 7
Q = (7, 7)
check: midpoint of (−3, 1) and (7, 7) is ((−3+7)/2, (1+7)/2) = (2, 4) ✓
WE 6Equation of the perpendicular bisector
The points A(1, 2) and B(7, 6) lie on a coordinate plane. Find the equation of the perpendicular bisector of [AB], giving your answer in the form y = mx + c.
Step 1: Find the midpoint of [AB]
M = ((1+7)/2, (2+6)/2) = (4, 4)
Step 2: Find the gradient of AB
mAB = (6−2)/(7−1) = 4/6 = 2/3
Step 3: Find the perpendicular gradient
mperp = −1/(2/3) = −3/2
Step 4: Use point-gradient form through M(4, 4)
y − 4 = −3/2 (x − 4)
y = −3x/2 + 6 + 4
y = −3x/2 + 10
the perpendicular bisector passes through the midpoint and is perpendicular to the original — uses all three formulas
💡 Top tips
- Label coordinates above the points before substituting. It takes two seconds and prevents almost every sign error.
- Bracket the negatives: writing 6 − (−2) is much safer than 6 − −2 and easier to read.
- For distance, leave the answer as an exact surd unless the question asks for a decimal — √50 is more precise than 7.07.
- Recognise Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25 — they show up surprisingly often in exam questions and let you spot the answer faster.
- Perpendicular gradient = negative reciprocal: flip the fraction and change the sign. m = 2/3 → mperp = −3/2.
- Use a sketch on every coordinate-geometry question. Sign errors are easy to catch visually but invisible in algebra.
- Check by substitution: plug the midpoint or perpendicular-bisector point back into your formulas to verify.
⚠ Common mistakes
- Subtracting in the wrong order for gradient. The first y in the numerator must pair with the first x in the denominator — otherwise you flip the sign.
- Forgetting to square the differences in the distance formula. d = √((Δx)² + (Δy)²), not √(Δx + Δy).
- Adding instead of averaging for the midpoint. (x1 + x2)/2, not just x1 + x2.
- Sign errors with negative coordinates. (−3) − (−5) = 2, not −8 or −2 or 8.
- Computing the reciprocal but forgetting to negate for perpendicular gradients. m = 2/3 gives 3/2 (reciprocal) — but the perpendicular is −3/2.
- Mistaking the gradient for the equation of the line. m = 4 means the gradient is 4 — not that the equation is y = 4.
- Writing distance with a ± sign. Distance is always non-negative — take the positive square root.
That’s the prior-learning toolkit you need before diving deeper into Topic 3. The next note, Arcs & Sectors Using Degrees, applies similar fraction-of-the-whole reasoning to circles — the arc is a fraction of the circumference, the sector a fraction of the area. Once you’ve got the angle measured in degrees, both formulas reduce to multiplying by θ/360.
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