Coordinate Geometry Notes

Cambridge IGCSE Maths: Revision notes

Key facts

Coordinates are written $(x, y)$ : across first, then up.
Length of a line: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ .
Midpoint: $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ , which adds the coordinates.
Three points are collinear if the gradients between pairs are equal.
For a missing endpoint, rearrange the midpoint formula.

Cartesian coordinates

An ordered pair (x, y) gives a point's horizontal and vertical position.

The $x$ -coordinate gives the horizontal position and the $y$ -coordinate the vertical position, always written $(x, y)$ . This lets us describe and calculate properties of lines and shapes algebraically.

The point (3, 2) is 3 across and 2 up

Where is the point (3, 2)?

Length of a line

Use Pythagoras: the horizontal and vertical distances are the two shorter sides of a right-angled triangle.

The line segment itself is the hypotenuse. Show the substitution into the formula as your method line: examiners award the method mark for correct substitution before the arithmetic is simplified.

Distance as the hypotenuse of a right-angled triangle

Length $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Worked example

Find the distance between $(1, 2)$ and $(4, 6)$ .

1
Substitute: $\sqrt{(4-1)^2+(6-2)^2}$ .
2
Simplify: $\sqrt{9+16} = \sqrt{25}$ .

Find the distance between (0, 0) and (6, 8).

Midpoint

The midpoint is the average of the two x-coordinates and the average of the two y-coordinates.

The midpoint formula always uses a sum, not a difference: subtracting before dividing by 2 is a common error.

The midpoint M of (2, -3) and (8, 5) is (5, 1)

Midpoint $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Worked example

Find the midpoint of $(2, -3)$ and $(8, 5)$ .

1
$x$ : $\frac{2+8}{2} = 5$ .
2
$y$ : $\frac{-3+5}{2} = 1$ .

Find the midpoint of (0, 4) and (6, 10).

Working without a diagram

Use the distance and midpoint formulas and gradients rather than relying on a drawing.

To check whether three points are collinear, calculate the gradients between pairs: if they are equal the points lie on a straight line. To find a missing coordinate from a midpoint, rearrange the midpoint formula.

A(1, 1), B(3, 5) and C(5, 9) all lie on y = 2x − 1

Distance

Use the length formula

Midpoint

Use the midpoint formula

Collinear

Equal gradients between pairs

Missing endpoint

Rearrange the midpoint formula

Worked example

$M(4, 1)$ is the midpoint of $A(2, -3)$ and $B$ . Find $B$ .

1
For $x$ : $\frac{2+x_B}{2}=4$ , so $x_B = 6$ .
2
For $y$ : $\frac{-3+y_B}{2}=1$ , so $y_B = 5$ .

A(1,1), B(3,5), C(5,9): gradient AB = 2. What is gradient BC?

Try an exam question

$A$ is the point $(-1, 2)$ and $B$ is the point $(5, 10)$ . (a) Find the midpoint of $AB$ . (b) Find the length of $AB$ .

[4 marks]

That's the notes covered.

Carry on to the next subtopic.