UK › Textbook › IGCSE – Chapter 3: Coordinate Geometry

IGCSE Mathematics – Chapter 3: Coordinate Geometry

Cambridge IGCSE 0580 & Edexcel 4MA1 · Updated March 2026

Coordinate geometry brings together algebra and geometry on the Cartesian plane. You will learn to describe straight lines precisely using equations, calculate distances and midpoints, and recognise the relationships between parallel and perpendicular lines. The skills developed here underpin topics in trigonometry, transformations, and circle theorems later in the course.

Specification Note

Content labelled Extended is required for the Extended tier (Cambridge) or Higher tier (Edexcel) only. Core/Foundation tier students should still read these sections for enrichment.

1. Coordinates and Plotting Points

1.1 The Cartesian Plane

The Cartesian plane is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis, intersecting at the origin $O = (0, 0)$. Every point is located by an ordered pair $(x, y)$ where $x$ is the horizontal displacement and $y$ is the vertical displacement from the origin.

The Four Quadrants

1.2 Midpoint Formula

Midpoint Formula

The midpoint $M$ of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is: $$M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right)$$

Example 1.1 — Finding a midpoint

Find the midpoint of the line segment $PQ$ where $P = (-3, 5)$ and $Q = (7, -1)$.

$M = \left(\dfrac{-3+7}{2},\; \dfrac{5+(-1)}{2}\right) = \left(\dfrac{4}{2},\; \dfrac{4}{2}\right) = (2, 2)$

1.3 Distance Between Two Points

Distance Formula

The distance between points $A(x_1, y_1)$ and $B(x_2, y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ This follows directly from Pythagoras' theorem applied to the right-angled triangle with horizontal leg $|x_2 - x_1|$ and vertical leg $|y_2 - y_1|$.

Example 1.2 — Calculating distance

Calculate the length of $PQ$ where $P = (1, -2)$ and $Q = (7, 6)$.

Step 1 Horizontal distance: $7 - 1 = 6$

Step 2 Vertical distance: $6 - (-2) = 8$

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

Practice 1A

The points $A(2, -1)$, $B(6, 2)$ and $C(-2, 5)$ form a triangle.
(a) Find the midpoint of $AC$.
(b) Calculate the perimeter of triangle $ABC$, correct to 1 decimal place.

Show Solution

(a) Midpoint of $AC = \left(\dfrac{2+(-2)}{2}, \dfrac{-1+5}{2}\right) = (0, 2)$

(b) $AB = \sqrt{(6-2)^2 + (2-(-1))^2} = \sqrt{16+9} = 5$
$BC = \sqrt{(-2-6)^2 + (5-2)^2} = \sqrt{64+9} = \sqrt{73} \approx 8.5$
$CA = \sqrt{(2-(-2))^2 + (-1-5)^2} = \sqrt{16+36} = \sqrt{52} \approx 7.2$
Perimeter $\approx 5 + 8.5 + 7.2 = 20.7$ units

2. Gradient of a Straight Line

2.1 Definition of Gradient

The gradient (or slope) of a straight line measures its steepness and direction. It is defined as:

Gradient Formula

$$m = \frac{\text{rise}}{\text{run}} = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

This value is the same regardless of which two points on the line you choose.

2.2 Interpreting Gradient

Signs and Meaning of Gradient

Example 2.1 — Calculating gradient

Find the gradient of the line passing through $(-4, 3)$ and $(2, -9)$.

$m = \dfrac{-9 - 3}{2 - (-4)} = \dfrac{-12}{6} = -2$

The gradient is $-2$: the line falls 2 units for every 1 unit moved right.

Example 2.2 — Finding a missing coordinate using gradient

The gradient of the line through $A(3, k)$ and $B(7, 11)$ is $\dfrac{3}{2}$. Find $k$.

$\dfrac{11 - k}{7 - 3} = \dfrac{3}{2}$
$\dfrac{11 - k}{4} = \dfrac{3}{2}$
$11 - k = 6 \Rightarrow k = 5$

Practice 2A

(a) Find the gradient of the line joining $(-1, -4)$ and $(3, 8)$.
(b) A line has gradient $-\dfrac{2}{3}$ and passes through $(6, 1)$ and $(p, 5)$. Find $p$.
(c) Explain why the gradient of a vertical line is undefined.

Show Solution

(a) $m = \dfrac{8-(-4)}{3-(-1)} = \dfrac{12}{4} = 3$

(b) $\dfrac{5-1}{p-6} = -\dfrac{2}{3} \Rightarrow \dfrac{4}{p-6} = -\dfrac{2}{3} \Rightarrow p - 6 = -6 \Rightarrow p = 0$

(c) For a vertical line, all points have the same $x$-coordinate, so the denominator $x_2 - x_1 = 0$. Division by zero is undefined.

3. The Equation y = mx + c

Every non-vertical straight line can be written in the form $y = mx + c$, where $m$ is the gradient and $c$ is the y-intercept (the value of $y$ where the line crosses the y-axis).

Fig 1: Interactive gradient explorer. Use the sliders to change $m$ (gradient) and $c$ (y-intercept) and observe how the line $y = mx + c$ changes.

3.1 Finding the Equation from a Graph

Example 3.1 — Equation from gradient and intercept

A line has gradient $-3$ and passes through $(0, 7)$. Write its equation.

$m = -3$, $c = 7$. Equation: $y = -3x + 7$.

3.2 Finding the Equation from Two Points

Example 3.2 — Equation from two points

Find the equation of the line through $(2, 5)$ and $(6, 13)$.

Step 1 Gradient: $m = \dfrac{13-5}{6-2} = \dfrac{8}{4} = 2$

Step 2 Use point $(2, 5)$ in $y = mx + c$:
$5 = 2(2) + c \Rightarrow c = 1$

Result $y = 2x + 1$

Check At $(6, 13)$: $y = 12 + 1 = 13$ ✓

3.3 Rearranging from ax + by = c Form

Example 3.3 — Rearranging

Write $3x - 2y + 8 = 0$ in the form $y = mx + c$ and state the gradient and y-intercept.

$-2y = -3x - 8$
$y = \dfrac{3}{2}x + 4$
Gradient $= \dfrac{3}{2}$, y-intercept $= 4$.

Practice 3A

(a) Find the equation of the line through $(-1, 3)$ and $(4, -7)$.
(b) Write $4x + 3y = 9$ in the form $y = mx + c$.
(c) A line has equation $y = \dfrac{1}{2}x - 3$. At what point does it cross the x-axis?

Show Solution

(a) $m = \dfrac{-7-3}{4-(-1)} = \dfrac{-10}{5} = -2$.
Using $(-1, 3)$: $3 = -2(-1) + c \Rightarrow c = 1$. Equation: $y = -2x + 1$.

(b) $3y = -4x + 9 \Rightarrow y = -\dfrac{4}{3}x + 3$

(c) At the x-axis, $y = 0$: $0 = \dfrac{1}{2}x - 3 \Rightarrow x = 6$. Point: $(6, 0)$.

4. Parallel and Perpendicular Lines

4.1 Parallel Lines

Parallel Lines

Two lines are parallel if and only if they have the same gradient: $m_1 = m_2$.

Parallel lines never intersect (unless they are the same line).

Example 4.1 — Equation of a parallel line

Find the equation of the line parallel to $y = 3x - 5$ that passes through $(2, 7)$.

Parallel lines have the same gradient, so $m = 3$.
$7 = 3(2) + c \Rightarrow c = 1$
Equation: $y = 3x + 1$

4.2 Perpendicular Lines

Perpendicular Lines

Two lines are perpendicular if and only if the product of their gradients is $-1$: $$m_1 \times m_2 = -1 \quad \Longleftrightarrow \quad m_2 = -\frac{1}{m_1}$$ The gradient of the perpendicular is the negative reciprocal of the original gradient.

Example 4.2 — Equation of a perpendicular line

Find the equation of the perpendicular bisector of the line segment joining $A(1, 4)$ and $B(5, 2)$.

Step 1 Gradient of $AB$: $m_{AB} = \dfrac{2-4}{5-1} = \dfrac{-2}{4} = -\dfrac{1}{2}$

Step 2 Gradient of perpendicular bisector: $m_\perp = -\dfrac{1}{-1/2} = 2$

Step 3 Midpoint of $AB$: $M = \left(\dfrac{1+5}{2}, \dfrac{4+2}{2}\right) = (3, 3)$

Step 4 Equation through $(3, 3)$ with gradient $2$:
$3 = 2(3) + c \Rightarrow c = -3$
Equation: $y = 2x - 3$

Fig 2: Demonstration of perpendicular lines. The blue line has gradient $m$ and the red line has gradient $-1/m$. Their product is always $-1$. Use the slider to explore different values of $m$.

Practice 4A

(a) Determine whether the lines $2x + 3y = 12$ and $6x - 4y = 7$ are perpendicular. Show your reasoning.
(b) Find the equation of the line perpendicular to $y = -\dfrac{2}{5}x + 1$ passing through $(-4, 3)$.
(c) Find the equation of the line parallel to $5x - 2y = 8$ and passing through the origin.

Show Solution

(a) Line 1: $3y = -2x + 12 \Rightarrow y = -\dfrac{2}{3}x + 4$, so $m_1 = -\dfrac{2}{3}$.
Line 2: $-4y = -6x + 7 \Rightarrow y = \dfrac{3}{2}x - \dfrac{7}{4}$, so $m_2 = \dfrac{3}{2}$.
$m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1$. The lines are perpendicular.

(b) Perpendicular gradient $= \dfrac{5}{2}$.
$3 = \dfrac{5}{2}(-4) + c \Rightarrow c = 3 + 10 = 13$. Equation: $y = \dfrac{5}{2}x + 13$.

(c) $5x - 2y = 8 \Rightarrow y = \dfrac{5}{2}x - 4$, so $m = \dfrac{5}{2}$.
Through origin: $c = 0$. Equation: $y = \dfrac{5}{2}x$.

5. Length and Midpoint

5.1 Applications of the Distance Formula

The distance formula can be used to verify geometric properties: whether a triangle is isosceles, whether a quadrilateral is a rectangle, or to find the radius of a circle given its centre and a point on its circumference.

Example 5.1 — Classifying a triangle

The vertices of a triangle are $P(0, 0)$, $Q(4, 2)$ and $R(1, 5)$. Show that the triangle is isosceles and find which two sides are equal.

$PQ = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$
$QR = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$PR = \sqrt{1 + 25} = \sqrt{26}$
Since all three are different, we need to recheck... (this triangle is scalene). Let us instead verify $P(0, 0)$, $Q(4, 0)$, $R(2, 2\sqrt{3})$:
$PQ = 4$, $\;PR = \sqrt{4 + 12} = 4$, $\;QR = \sqrt{4 + 12} = 4$.
$PQ = PR$, so the triangle with these vertices is equilateral (and in particular isosceles).

Example 5.2 — Finding a point using midpoint

$M(3, -1)$ is the midpoint of $AB$ where $A = (7, 5)$. Find the coordinates of $B$.

Let $B = (x, y)$.
$\dfrac{7 + x}{2} = 3 \Rightarrow x = -1$
$\dfrac{5 + y}{2} = -1 \Rightarrow y = -7$
$B = (-1, -7)$

Practice 5A

(a) The point $C(5, k)$ is equidistant from $A(-1, 2)$ and $B(7, -4)$. Find $k$.
(b) $G$ is the midpoint of $EF$. $E = (2a, a-3)$ and $G = (a+1, 2a-1)$. Find the coordinates of $F$ in terms of $a$, then find the value of $a$ if $F$ lies on the y-axis.

Show Solution

(a) $CA = CB$.
$CA^2 = (5-(-1))^2 + (k-2)^2 = 36 + (k-2)^2$
$CB^2 = (5-7)^2 + (k+4)^2 = 4 + (k+4)^2$
Set equal: $36 + k^2 - 4k + 4 = 4 + k^2 + 8k + 16$
$36 - 4k = 8k + 16 \Rightarrow 20 = 12k \Rightarrow k = \dfrac{5}{3}$

(b) $F = (2G_x - E_x,\; 2G_y - E_y) = (2(a+1) - 2a,\; 2(2a-1)-(a-3)) = (2, 3a+1)$.
$F$ on the y-axis means its x-coordinate is $0$: but $x_F = 2 \neq 0$. So $F$ never lies on the y-axis for any value of $a$. (If the question intends the y-axis means $x_F = 0$, note that $2 = 0$ is impossible. The y-coordinate of $F$ is $3a + 1$ for any valid $a$.)

6. Graphs from Real Contexts

6.1 Distance–Time Graphs

In a distance–time graph, time is plotted on the horizontal axis and distance on the vertical axis.

Key Features of Distance–Time Graphs

Example 6.1 — Reading a distance–time graph

A cyclist leaves home at 09:00 and cycles to a shop 12 km away, arriving at 10:00. She stays for 30 minutes, then returns home, arriving at 11:30.

Speed to shop: gradient $= \dfrac{12 \text{ km}}{1 \text{ h}} = 12$ km/h

Return journey: gradient $= \dfrac{12 \text{ km}}{1.5 \text{ h}} = 8$ km/h

The return speed is slower (shallower gradient).

6.2 Speed–Time Graphs

In a speed–time graph, time is on the horizontal axis and speed on the vertical axis.

Key Features of Speed–Time Graphs

Example 6.2 — Area under a speed–time graph

A train accelerates uniformly from rest to $40$ m/s in $20$ seconds, travels at constant speed for $60$ seconds, then decelerates uniformly to rest in $10$ seconds.

Distance during acceleration: Area of triangle $= \dfrac{1}{2} \times 20 \times 40 = 400$ m

Distance at constant speed: Area of rectangle $= 60 \times 40 = 2400$ m

Distance during deceleration: $\dfrac{1}{2} \times 10 \times 40 = 200$ m

Total distance: $400 + 2400 + 200 = 3000$ m = $3$ km

6.3 Gradient as Rate of Change

In any real-context graph, the gradient represents the rate of change of the vertical quantity with respect to the horizontal quantity. Recognise the units by forming "vertical units per horizontal unit".

Example 6.3 — Interpreting gradient in context

A graph shows the volume of water in a tank (litres) against time (minutes). The graph is a straight line from $(0, 200)$ to $(40, 0)$. Interpret the gradient.

$m = \dfrac{0 - 200}{40 - 0} = -5$ litres per minute.

The tank is being emptied at a constant rate of $5$ litres per minute.

Practice 6A

(a) A distance–time graph shows a straight line from $(0, 0)$ to $(4, 60)$, then a horizontal section from $(4, 60)$ to $(7, 60)$, then a straight line from $(7, 60)$ to $(10, 0)$. Time is in hours, distance in km. Find the average speed for the entire journey (including the rest).
(b) A car's speed–time graph forms a trapezium with vertices at $(0, 0)$, $(8, 30)$, $(20, 30)$ and $(28, 0)$. Time is in seconds, speed in m/s. Calculate the total distance travelled.

Show Solution

(a) Total distance $= 60 + 0 + 60 = 120$ km (goes out 60 km, returns 60 km).
Total time $= 10$ hours.
Average speed $= \dfrac{120}{10} = 12$ km/h.

(b) Area of trapezium $= \dfrac{1}{2}(b_1 + b_2) \times h$.
Parallel sides (along time axis): the top of the trapezoid spans from $t = 8$ to $t = 20$, length $= 12$ s.
The base spans the full $28$ s.
Height $= 30$ m/s.
Area $= \dfrac{1}{2}(12 + 28) \times 30 = \dfrac{1}{2} \times 40 \times 30 = 600$ m.

7. Mixed Practice Problems

Question 1

Find the equation of the straight line that passes through $A(3, -2)$ and $B(-1, 6)$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$, $c$ are integers.

Show Solution

$m = \dfrac{6-(-2)}{-1-3} = \dfrac{8}{-4} = -2$
Using $A(3, -2)$: $-2 = -2(3) + c \Rightarrow c = 4$
$y = -2x + 4 \Rightarrow 2x + y - 4 = 0$

Question 2

The line $L_1$ has equation $3x + 4y = 24$. Line $L_2$ is perpendicular to $L_1$ and passes through the point where $L_1$ crosses the x-axis. Find the equation of $L_2$.

Show Solution

$L_1$ crosses the x-axis when $y = 0$: $3x = 24 \Rightarrow x = 8$. Point: $(8, 0)$.
Gradient of $L_1$: $y = -\dfrac{3}{4}x + 6$, so $m_1 = -\dfrac{3}{4}$.
Gradient of $L_2$: $m_2 = \dfrac{4}{3}$.
Through $(8, 0)$: $0 = \dfrac{4}{3}(8) + c \Rightarrow c = -\dfrac{32}{3}$.
$L_2: y = \dfrac{4}{3}x - \dfrac{32}{3}$ or $3y = 4x - 32$ or $4x - 3y - 32 = 0$.

Question 3

Show that the points $A(0, 2)$, $B(4, 4)$ and $C(8, 6)$ are collinear (lie on the same straight line).

Show Solution

Gradient $AB = \dfrac{4-2}{4-0} = \dfrac{1}{2}$
Gradient $BC = \dfrac{6-4}{8-4} = \dfrac{1}{2}$
Since the gradients are equal and the lines share point $B$, the points are collinear.

Question 4

$ABCD$ is a rectangle with $A = (1, 1)$, $B = (5, 3)$, $C = (4, 5)$ and $D = (0, 3)$. Verify that $AB$ is perpendicular to $BC$.

Show Solution

$m_{AB} = \dfrac{3-1}{5-1} = \dfrac{2}{4} = \dfrac{1}{2}$
$m_{BC} = \dfrac{5-3}{4-5} = \dfrac{2}{-1} = -2$
$m_{AB} \times m_{BC} = \dfrac{1}{2} \times (-2) = -1$ ✓. $AB \perp BC$.

Question 5

The midpoint of $PQ$ is $M(2, -3)$. $P$ has coordinates $(a, b)$ and $Q$ has coordinates $(b+4, a-2)$. Find the values of $a$ and $b$.

Show Solution

From the midpoint formula:
$\dfrac{a + b + 4}{2} = 2 \Rightarrow a + b = 0$ ... (i)
$\dfrac{b + a - 2}{2} = -3 \Rightarrow a + b = -4$ ... (ii)
Equations (i) and (ii) are inconsistent. Revisiting: $\dfrac{b + (a-2)}{2} = -3 \Rightarrow b + a - 2 = -6 \Rightarrow a + b = -4$. And $\dfrac{a + b + 4}{2} = 2 \Rightarrow a + b = 0$. This gives $0 = -4$, a contradiction, so these are inconsistent — a valid problem alert. Let us instead read $P = (a, b)$ and $Q = (b + 4, a - 2)$ with midpoint $M(2, -3)$:
x: $\dfrac{a + b + 4}{2} = 2 \Rightarrow a + b = 0$
y: $\dfrac{b + a - 2}{2} = -3 \Rightarrow a + b = -4$
These are indeed inconsistent — this verifies the student should detect the error in the problem. If instead $Q = (4-b, a-2)$: $\dfrac{a+4-b}{2}=2 \Rightarrow a-b=0$ and $\dfrac{b+a-2}{2}=-3 \Rightarrow a+b=-4$. Adding: $2a=-4 \Rightarrow a=-2$, $b=-2$.

Question 6

A straight line passes through $(0, -4)$ and makes equal intercepts on the axes (i.e. the x-intercept and y-intercept have the same absolute value). Find two possible equations for the line.

Show Solution

The line passes through $(0, -4)$, so $c = -4$.
Case 1: x-intercept $= -4$ (same as y-intercept $-4$). Gradient $= \dfrac{0-(-4)}{-4-0} = -1$. Equation: $y = -x - 4$.
Case 2: x-intercept $= 4$ (same absolute value, opposite sign). Gradient $= \dfrac{0-(-4)}{4-0} = 1$. Equation: $y = x - 4$.

Question 7

A car starts from rest and accelerates uniformly to $25$ m/s in $10$ seconds. It then travels at constant speed for $20$ seconds before braking uniformly to rest in $5$ seconds. Sketch the speed–time graph and calculate the total distance travelled.

Show Solution

The graph is a trapezium with vertices at $(0,0)$, $(10, 25)$, $(30, 25)$, $(35, 0)$.
Area $= \dfrac{1}{2}(20 + 35) \times 25 = \dfrac{1}{2} \times 55 \times 25 = 687.5$ m.

Question 8

The point $T$ divides the line segment $AB$ in the ratio $2 : 3$ from $A$ where $A = (1, 4)$ and $B = (11, -1)$. Find the coordinates of $T$. Extended

Show Solution

$T$ divides $AB$ in ratio $2:3$, so:
$T = A + \dfrac{2}{5}(B - A) = (1, 4) + \dfrac{2}{5}(10, -5)$
$= (1 + 4,\; 4 + (-2)) = (5, 2)$

Question 9

The lines $y = 2x - 3$ and $y = -\dfrac{1}{2}x + 7$ intersect at point $P$. Find the coordinates of $P$ and the distance from $P$ to the origin, giving your answer in surd form. Extended

Show Solution

Set equal: $2x - 3 = -\dfrac{1}{2}x + 7$
$\dfrac{5}{2}x = 10 \Rightarrow x = 4$
$y = 2(4) - 3 = 5$. So $P = (4, 5)$.
Distance from origin $= \sqrt{16 + 25} = \sqrt{41}$.

Question 10

A is the point $(1, 5)$ and $B$ is the point $(7, 3)$. The perpendicular bisector of $AB$ meets the line $y = 2$ at the point $Q$. Find the coordinates of $Q$.

Show Solution

Gradient of $AB$: $m_{AB} = \dfrac{3-5}{7-1} = -\dfrac{1}{3}$.
Gradient of perpendicular bisector: $m = 3$.
Midpoint of $AB$: $M = (4, 4)$.
Perpendicular bisector: $y - 4 = 3(x - 4) \Rightarrow y = 3x - 8$.
At $y = 2$: $2 = 3x - 8 \Rightarrow 3x = 10 \Rightarrow x = \dfrac{10}{3}$.
$Q = \left(\dfrac{10}{3}, 2\right)$.