Coordinate GeometryEdexcel IGCSE Maths: Revision notes

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the distance formula comes straight from Pythagoras' theorem: the horizontal and vertical gaps between the points form the two shorter sides of a right-angled triangle, and the line $AB$ is the hypotenuse.

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

Worked example: Find the distance between $A(1, 2)$ and $B(4, 6)$.

$d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Always subtract in the same order for both coordinates — it doesn't matter which point you call $A$ or $B$ because the differences get squared anyway.

The midpoint is the point exactly halfway between two coordinates. Average the $x$-coordinates and average the $y$-coordinates separately.

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Worked example: Find the midpoint of $A(1, 2)$ and $B(4, 6)$.

$M = \left(\frac{1+4}{2}, \frac{2+6}{2}\right) = \left(\frac{5}{2}, 4\right) = (2.5, 4)$

This also works in reverse: if you know one endpoint and the midpoint, you can find the missing endpoint by rearranging.

To find the point $P$ that divides the segment from $A(x_1, y_1)$ to $B(x_2, y_2)$ in the ratio $m : n$ (measured from $A$ to $B$), use the section formula:

$P = \left(\frac{n x_1 + m x_2}{m+n}, \frac{n y_1 + m y_2}{m+n}\right)$

Worked example: Find the point $P$ that divides $A(0, 0)$ to $B(10, 15)$ in the ratio $2:3$.

Here $m = 2$, $n = 3$, so $P$ is $\frac{2}{5}$ of the way from $A$ to $B$:

$P = \left(0 + \frac{2}{5}(10-0),\\; 0 + \frac{2}{5}(15-0)\right) = (4, 6)$

A quicker way many students prefer: work out the total number of parts ($m+n$), find $\frac{1}{m+n}$ of the total change in $x$ and $y$, then step across from $A$ by $m$ parts.

Exam questions often chain coordinate geometry with straight-line graphs. For example, you might be asked to find the midpoint of a diameter to get the centre of a circle, then use the distance formula to find the radius. Or you might use the section formula to find a point that divides a segment in a given ratio, then find the gradient of a line through that point.

Always sketch a quick diagram first — even a rough one helps you spot whether your answer is sensible (e.g. the midpoint should look roughly halfway between the two plotted points).