Worked example 1: Gradient between two points
Find the gradient of the line through $A(-3, 7)$ and $B(5, -9)$.
Solution
1. Formula: $m = \frac{y_2-y_1}{x_2-x_1}$.
2. Substitute: $m = \frac{-9-7}{5-(-3)} = \frac{-16}{8}$.
3. State: $\mathbf{m = -2}$.
IB Maths AI SL · Unit 3: Geometry and Trigonometry
Exam-style IB Maths AI SL coordinate geometry questions with worked solutions. Start with the 5 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
SL AI's geometry is applied: which cereal box has the lowest surface-area-to-volume ratio? Master the volume + surface-area formulas for cylinders, cones, and truncated shapes. 3D geometry — volume, surface area, and packaging overview →
The question bank covers these coordinate geometry question types (number of questions in brackets):
Find the gradient of the line through $A(-3, 7)$ and $B(5, -9)$.
1. Formula: $m = \frac{y_2-y_1}{x_2-x_1}$.
2. Substitute: $m = \frac{-9-7}{5-(-3)} = \frac{-16}{8}$.
3. State: $\mathbf{m = -2}$.
A line has gradient $3$ and passes through $(2, 5)$. Find its equation in $y=mx+c$ form.
1. Point-slope: $y - 5 = 3(x - 2)$.
2. Expand: $y - 5 = 3x - 6$.
3. Rearrange: $\mathbf{y = 3x - 1}$.
$L_1$: $4x + 2y = 10$. $L_2$ is parallel to $L_1$ and passes through $(1, 6)$. Find the x-intercept of $L_2$.
1. Rearrange $L_1$: $y = -2x + 5$, gradient $-2$.
2. $L_2$: $y - 6 = -2(x - 1) \implies y = -2x + 8$.
3. Set $y = 0$: $x = 4$.
4. State: $\mathbf{(4, 0)}$.
Line $L$: $3x - 5y = 15$. Find the perpendicular gradient.
1. Rearrange: $y = \frac{3}{5}x - 3$, $m_1 = \frac{3}{5}$.
2. Perpendicular: negative reciprocal of $\frac{3}{5}$.
3. State: $\mathbf{m_2 = -\frac{5}{3}}$.
Towns at $A(2, 4)$ and $B(8, 16)$. Find the equidistant boundary line in $y=mx+c$ form.
1. Midpoint: $M = (5, 10)$.
2. Gradient of $AB$: $\frac{12}{6} = 2$.
3. Perpendicular gradient: $-0.5$.
4. Line through $M$: $y - 10 = -0.5(x - 5)$.
5. State: $\mathbf{y = -0.5x + 12.5}$.
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
The distance between two points with coordinates \((x_1, y_1)\) and \((x_2, y_2)\) is equal to \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Consider the points \(A(1, -2)\) and \(B(40, -100)\).
Calculate the exact distance between points A and B.
Find the exact coordinates of the midpoint of the line segment [AB].
Dilara is designing a kite ABCD on a coordinate grid. The coordinates of A, B, and C are \(A(2, 0)\), \(B(0, 4)\), and \(C(4, 6)\) respectively. Point D lies on the \(x\)-axis. The diagonals [AC] and [BD] are perpendicular.
Find the gradient of the line through A and C.
Write down the gradient of the line through B and D.
Find the equation of the line through B and D.
Hence, write down the \(x\)-coordinate of point D.
The equation of a straight coastline is modelled by the line \(L_1: 2y - x - 10 = 0\). A boat is anchored at point \(M(8, 18)\). The coastguard needs to find the shortest distance from the boat to the coastline.
Find the gradient of the coastline \(L_1\).
Find the equation of the line \(L_2\), which passes through \(M\) and is perpendicular to \(L_1\).
Find the coordinates of point \(D\), the intersection of \(L_1\) and \(L_2\).
Calculate the shortest distance from the boat to the coastline.
There are 32 exam-style coordinate geometry questions in the AI SL question bank (Paper 1: 19 · Paper 2: 13), graded 7 easy, 8 medium, 8 hard, 7 very hard, 2 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Both. In the bank, Paper 1: 19 · Paper 2: 13. Practise with your GDC — AI papers expect calculator methods throughout.
Open the AI SL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.