Coordinate geometry turns geometry into algebra. Once every point has coordinates, slopes, lengths and midpoints are found by calculation, not by measuring. It is examined every year in O-Level E-Math Paper 2, and it is the prerequisite for A-Math coordinate geometry, equations of circles and linear law.
Four formulas carry the whole topic: gradient, length, midpoint and the point-gradient form. Every later question, from parallel lines and collinearity to finding unknown coordinates and areas, is built from these four, so they need to become automatic.
What these notes cover
- •Gradient of a straight line
- •Length of a line segment
- •Midpoint and working backwards
- •The equation y = mx + c
- •Point-gradient form
- •Parallel lines
- •Collinear points
- •Area of a rectilinear figure
- •Proving quadrilaterals with coordinates
- •Fourth vertex of a parallelogram
- •Common mistakes and exam technique
Key definitions
- Gradient
- How steep a line is: how much y changes when x increases by 1 unit. It is rise over run, the vertical change divided by the horizontal change.
- Midpoint
- The point halfway along a line segment, found by averaging the x-coordinates and averaging the y-coordinates of the two ends.
- Parallel lines
- Lines that are equally steep and never meet. They have equal gradients and can differ only in their y-intercept.
- Collinear points
- Three points are collinear if they lie on the same straight line.
- Rectilinear figure
- Any shape bounded by straight lines, such as a triangle, quadrilateral or pentagon.
Key formulas
Gradient of AB
For A(x₁, y₁) and B(x₂, y₂). Either point can be point 1, but the same point must come first on the top and on the bottom. Mixing the order flips the sign.
Length of AB
From Pythagoras' theorem. Squaring removes negative signs, so the order of subtraction does not matter here. Give an exact surd, simplified where possible (e.g. √80 = 4√5), unless 3 s.f. is asked for.
Midpoint of AB
The average of the coordinates, so add, do not subtract. To find the other end B from the midpoint M and end A, double M and subtract A: x₂ = 2 × (x-coordinate of M) − x₁, and likewise for y.
Equation of a straight line
y = mx + c
m is the gradient and c is the y-intercept, where the line cuts the y-axis at (0, c). Rearrange into the form y = … before reading off m and c. A vertical line x = k cannot be written in this form.
Point-gradient form
y − y₁ = m(x − x₁)
The line with gradient m through the point (x₁, y₁). Expand and rearrange into y = mx + c or the form asked for, then check by substituting another point.
Fourth vertex of parallelogram ABCD
D = A + C − B
Comes from the diagonals bisecting each other (sharing a midpoint). Apply it separately to the x- and y-coordinates. A and C must be opposite vertices. Also works for rectangles, rhombuses and squares.
Worked example
ABCD is a rectangle with A(2, 1) and C(7, 14). Find (a) the gradient of AC, (b) the length of AC, (c) the coordinates of the point where the diagonals meet, (d) the equation of AC.
- 1(a) Gradient of AC: m = (14 − 1) ÷ (7 − 2) = 13 ÷ 5.
- 2(b) Length of AC: AC = √[(7 − 2)² + (14 − 1)²] = √(25 + 169) = √194 ≈ 13.9 units (3 s.f.).
- 3(c) The diagonals of a rectangle bisect each other, so they meet at the midpoint of AC: M = ((2 + 7) ÷ 2, (1 + 14) ÷ 2) = (4.5, 7.5).
- 4(d) Using A(2, 1) and m = 13 ÷ 5: y − 1 = (13 ÷ 5)(x − 2), so y = (13 ÷ 5)x − (26 ÷ 5) + (5 ÷ 5) = (13 ÷ 5)x − (21 ÷ 5), which can also be written as 5y = 13x − 21.
- 5Check with C(7, 14): (91 ÷ 5) − (21 ÷ 5) = 70 ÷ 5 = 14. Correct.
Watch out: In part (c), state the geometric reason (the diagonals bisect each other). It shows the examiner why a midpoint is the right tool.
Common mistakes
Inconsistent order in the gradient
Wrong: m = (y₂ − y₁) ÷ (x₁ − x₂)
Right: m = (y₂ − y₁) ÷ (x₂ − x₁). Use the same point first on the top and the bottom; mixing the order flips the sign of the gradient.
Reading m before rearranging
Wrong: For 2y + 6x = 5, m = 6.
Right: Make y the subject first: 2y = −6x + 5, so y = −3x + (5 ÷ 2). Hence m = −3 and c = 5 ÷ 2.
Missing the common point in a collinearity proof
Wrong: Gradient of AB = gradient of CD, so A, B, C, D are collinear.
Right: Equal gradients alone only show the lines are parallel. Show gradient of AB = gradient of BC and state that B is a common point.
Questions students ask
How do you find the gradient of a line from two points?
Use m = (y₂ − y₁) ÷ (x₂ − x₁), the vertical change divided by the horizontal change. Either point can be point 1, as long as the same point comes first on the top and the bottom. Put brackets round negative coordinates, e.g. 4 − (−1) = 5.
How do you prove three points are collinear?
Show that the gradient of AB equals the gradient of BC, then state that B is a common point. Equal gradients on their own only show the lines are parallel; because AB and BC share B, they must be one straight line.
How do you find the area of a triangle on a coordinate grid?
If a side is horizontal or vertical, use it as the base and find ½ × base × height. Otherwise, enclose the figure in the smallest rectangle with horizontal and vertical sides and subtract the right-angled triangles (and trapezia) outside it. Every length needed is found by subtracting coordinates, so no square roots are needed.
How do you find the fourth vertex of a parallelogram?
The diagonals of a parallelogram bisect each other, so for ABCD the shortcut is D = A + C − B, applied to the x- and y-coordinates separately. A and C must be opposite vertices. It also works for rectangles, rhombuses and squares.
In the full PDF
- •Diagram of the rectangle on axes for the worked example
- •Further worked examples: finding an unknown coordinate, a length that gives two answers, a parallel line through a point and collinearity with an unknown
- •Area worked examples using enclose and subtract, including proving a rhombus and finding its area
- •Full table of coordinate tests for parallelograms, rhombuses, rectangles, squares, trapezia and triangles
- •Table of gradient signs: positive, negative, zero and undefined
- •Remaining common mistakes, an exam-technique checklist and section summaries