Coordinate Geometry – GCSE Maths

Nearly every coordinate geometry question uses one of these three, applied to two points $(x_1,\ y_1)$ and $(x_2,\ y_2)$.

Midpoint

$M = \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)$ — average the coordinates

Length (distance)

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

Gradient

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ — rise over run

Worked Example 1 — All three at once

$A = (1,\ 2)$ and $B = (7,\ 10)$. Find the midpoint, the length $AB$ and the gradient of $AB$.

①Midpoint: $\left(\dfrac{1+7}{2},\ \dfrac{2+10}{2}\right) = (4,\ 6)$

②Length: change in $x = 6$, change in $y = 8$.

③$AB = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

④Gradient: $m = \dfrac{10 - 2}{7 - 1} = \dfrac{8}{6} = \dfrac{4}{3}$

Subtract in the same order. For the gradient, if you do $y_2 - y_1$ on the top you must do $x_2 - x_1$ on the bottom. Mixing the order flips the sign.

Straight-line equation

$y = mx + c$ — $m$ is the gradient, $c$ is the $y$-intercept

Find the gradient $m$ using two points.
Substitute $m$ and the coordinates of one point into $y = mx + c$.
Solve for $c$.
Write out the full equation.

Worked Example 2 — Line through two points

Find the equation of the line through $(2,\ 5)$ and $(6,\ 17)$.

①$m = \dfrac{17 - 5}{6 - 2} = \dfrac{12}{4} = 3$

②Substitute $(2,\ 5)$: $\;5 = 3(2) + c$

③$5 = 6 + c \Rightarrow c = -1$

④$y = 3x - 1$

Check with the other point: $3(6) - 1 = 17$ ✓

The two rules

Parallel: the gradients are equal, $m_1 = m_2$
Perpendicular: the gradients multiply to $-1$, $m_1 \times m_2 = -1$

Finding a perpendicular gradient quickly: turn the fraction upside down and change the sign — the negative reciprocal.
$m = 3 = \tfrac{3}{1} \Rightarrow$ perpendicular gradient $= -\tfrac{1}{3}$
$m = -\tfrac{2}{5} \Rightarrow$ perpendicular gradient $= \tfrac{5}{2}$

Worked Example 3 — Perpendicular line

Find the equation of the line perpendicular to $y = 4x - 3$ that passes through $(8,\ 1)$.

①The given gradient is $4$.

②Perpendicular gradient $= -\dfrac{1}{4}$

③Substitute $(8,\ 1)$: $\;1 = -\tfrac{1}{4}(8) + c = -2 + c$

④$c = 3$

⑤$y = -\dfrac{1}{4}x + 3$

Worked Example 4 — Perpendicular bisector

Find the equation of the perpendicular bisector of the line joining $A(1,\ 3)$ and $B(7,\ 11)$.

①Midpoint: $\left(\dfrac{1+7}{2},\ \dfrac{3+11}{2}\right) = (4,\ 7)$

②Gradient of $AB$: $\dfrac{11-3}{7-1} = \dfrac{8}{6} = \dfrac{4}{3}$

③Perpendicular gradient: $-\dfrac{3}{4}$

④Through $(4,\ 7)$: $\;7 = -\tfrac{3}{4}(4) + c = -3 + c$, so $c = 10$.

⑤$y = -\dfrac{3}{4}x + 10$

Rearrange first. If a line is given as $2x + 3y = 12$, you cannot read the gradient off directly. Rearrange to $y = -\tfrac{2}{3}x + 4$, so $m = -\tfrac{2}{3}$.

To prove what kind of shape four points form, calculate lengths and gradients and compare them.

To prove…	Show that…
Two sides are parallel	Their gradients are equal
Two sides are perpendicular	Their gradients multiply to $-1$
Two sides are equal	Their lengths are equal
It is a parallelogram	Both pairs of opposite sides are parallel
It is a rhombus	All four sides are equal
It is a rectangle	Opposite sides parallel and one angle is $90^\circ$
It is a square	All sides equal and one angle is $90^\circ$
A triangle is right-angled	Two sides have gradients multiplying to $-1$
A triangle is isosceles	Two sides have equal length
Three points are collinear	The gradient between the first two equals that between the last two

Worked Example 5 — Proving a shape is a parallelogram

$A(1,\ 1)$, $B(4,\ 3)$, $C(9,\ 3)$ and $D(6,\ 1)$. Show that $ABCD$ is a parallelogram.

①Gradient $AB = \dfrac{3-1}{4-1} = \dfrac{2}{3}$

②Gradient $DC = \dfrac{3-1}{9-6} = \dfrac{2}{3}$ — equal, so $AB \parallel DC$

③Gradient $BC = \dfrac{3-3}{9-4} = 0$

④Gradient $AD = \dfrac{1-1}{6-1} = 0$ — equal, so $BC \parallel AD$

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram. ∎

Worked Example 6 — Right-angled triangle

Show that the triangle with vertices $P(0,\ 2)$, $Q(4,\ 4)$ and $R(2,\ 8)$ is right-angled.

①Gradient $PQ = \dfrac{4-2}{4-0} = \dfrac{2}{4} = \dfrac{1}{2}$

②Gradient $QR = \dfrac{8-4}{2-4} = \dfrac{4}{-2} = -2$

③$\dfrac{1}{2} \times (-2) = -1$

The product is $-1$, so $PQ \perp QR$ and the triangle is right-angled at $Q$. ∎

Alternative: use Pythagoras. $PQ^2 = 20$, $QR^2 = 20$, $PR^2 = 4 + 36 = 40$. Since $20 + 20 = 40$, the triangle is right-angled — and also isosceles.

Point dividing $AB$ in the ratio $m : n$

Move $\dfrac{m}{m+n}$ of the way from $A$ to $B$

Worked Example 7 — Ratio point

$A = (2,\ 1)$ and $B = (14,\ 13)$. Find the point $P$ with $AP : PB = 1 : 3$.

①Total parts $= 4$, so $P$ is $\tfrac{1}{4}$ of the way from $A$ to $B$.

②Change in $x = 14 - 2 = 12$; a quarter is $3$.

③Change in $y = 13 - 1 = 12$; a quarter is $3$.

④$P = (2 + 3,\ 1 + 3) = (5,\ 4)$

Midpoint

Average both coordinates.

Length

$\sqrt{(\Delta x)^2 + (\Delta y)^2}$ — Pythagoras.

Gradient

$\dfrac{\Delta y}{\Delta x}$, subtracting in the same order both times.

Line equation

$y = mx + c$; find $m$, then substitute a point to get $c$.

Parallel

Equal gradients.

Perpendicular

Negative reciprocal; the product of the gradients is $-1$.

Rearrange

Get the equation into $y = mx + c$ before reading off $m$.

Proving shapes

Compare lengths for equal sides, gradients for parallel and perpendicular.

Collinear

Same gradient between successive pairs of points.

Question 1

Find the midpoint of $A(-3,\ 7)$ and $B(5,\ -1)$.

▶ Show solution

$\left(\dfrac{-3+5}{2},\ \dfrac{7+(-1)}{2}\right) = \left(\dfrac{2}{2},\ \dfrac{6}{2}\right) = (1,\ 3)$

Question 2

Find the exact length of the line joining $(2,\ 3)$ and $(7,\ 15)$.

▶ Show solution

$\Delta x = 5$, $\Delta y = 12$

$d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Question 3

Find the gradient of the line through $(-2,\ 8)$ and $(4,\ -4)$.

▶ Show solution

$m = \dfrac{-4 - 8}{4 - (-2)} = \dfrac{-12}{6} = -2$

Question 4

Find the equation of the line through $(1,\ -2)$ and $(5,\ 6)$.

▶ Show solution

$m = \dfrac{6 - (-2)}{5 - 1} = \dfrac{8}{4} = 2$

Substitute $(1,\ -2)$: $-2 = 2(1) + c$, so $c = -4$.

$y = 2x - 4$

Question 5

A line has equation $3x + 2y = 18$. Find (a) its gradient, (b) the gradient of a perpendicular line.

▶ Show solution

(a) Rearrange: $2y = -3x + 18$, so $y = -\tfrac{3}{2}x + 9$. Gradient $= -\dfrac{3}{2}$.

(b) Negative reciprocal: $\dfrac{2}{3}$.

Check: $-\tfrac{3}{2} \times \tfrac{2}{3} = -1$ ✓

Question 6

Find the equation of the line parallel to $y = 5x + 2$ passing through $(3,\ 4)$.

▶ Show solution

Parallel lines have the same gradient, so $m = 5$.

$4 = 5(3) + c = 15 + c$, so $c = -11$.

$y = 5x - 11$

Question 7

Show that $A(1,\ 2)$, $B(3,\ 6)$ and $C(6,\ 12)$ are collinear.

▶ Show solution

Gradient $AB = \dfrac{6-2}{3-1} = \dfrac{4}{2} = 2$

Gradient $BC = \dfrac{12-6}{6-3} = \dfrac{6}{3} = 2$

The gradients are equal and the lines share the point $B$, so all three points lie on the same straight line. ∎

Question 8

Find the equation of the perpendicular bisector of the line joining $(2,\ 1)$ and $(8,\ 5)$.

▶ Show solution

Midpoint: $\left(\dfrac{2+8}{2},\ \dfrac{1+5}{2}\right) = (5,\ 3)$

Gradient of the segment: $\dfrac{5-1}{8-2} = \dfrac{4}{6} = \dfrac{2}{3}$

Perpendicular gradient: $-\dfrac{3}{2}$

$3 = -\tfrac{3}{2}(5) + c = -7.5 + c$, so $c = 10.5$.

$y = -\dfrac{3}{2}x + 10.5$

Question 9

$P(0,\ 0)$, $Q(4,\ 2)$, $R(6,\ 6)$ and $S(2,\ 4)$. Show that $PQRS$ is a rhombus.

▶ Show solution

$PQ = \sqrt{4^2 + 2^2} = \sqrt{20}$

$QR = \sqrt{2^2 + 4^2} = \sqrt{20}$

$RS = \sqrt{(-4)^2 + (-2)^2} = \sqrt{20}$

$SP = \sqrt{(-2)^2 + (-4)^2} = \sqrt{20}$

All four sides are equal, so $PQRS$ is a rhombus.

Not a square: gradient $PQ = \tfrac{1}{2}$ and gradient $QR = 2$; the product is $1$, not $-1$, so the angle at $Q$ is not $90^\circ$. ∎

Question 10

The line $L$ has equation $y = 2x - 6$. Point $A$ is $(1,\ 7)$.

(a) Find the equation of the line through $A$ perpendicular to $L$. (b) Find the point $B$ where the two lines cross. (c) Find the shortest distance from $A$ to $L$, in exact form.

▶ Show solution

(a) Gradient of $L$ is $2$, so the perpendicular gradient is $-\tfrac{1}{2}$.

$7 = -\tfrac{1}{2}(1) + c$, so $c = 7.5$.

Equation: $y = -\tfrac{1}{2}x + 7.5$

(b) Set the two equal: $2x - 6 = -\tfrac{1}{2}x + 7.5$

$2.5x = 13.5 \Rightarrow x = 5.4$

$y = 2(5.4) - 6 = 4.8$, so $B = (5.4,\ 4.8)$.

(c) The shortest distance is $AB$ (perpendicular distance).

$\Delta x = 4.4$, $\Delta y = -2.2$

$AB = \sqrt{4.4^2 + 2.2^2} = \sqrt{19.36 + 4.84} = \sqrt{24.2}$

$= \sqrt{24.2} = 4.92$ (3 s.f.), or exactly $\dfrac{11\sqrt{5}}{5}$.