Coordinate geometry for IGCSE Mathematics: Methods and Worked Examples

The principle behind coordinate geometry

Every point on a flat surface can be described by exactly two numbers: a horizontal position and a vertical position. That principle, first formalised by Rene Descartes in the seventeenth century, turns geometric problems into algebraic ones. Instead of reasoning about shapes through diagrams alone, you gain the power to calculate distances, gradients, midpoints, and equations with precision. For IGCSE Mathematics, coordinate geometry bridges the gap between algebra and geometry, and the examiners test that bridge frequently.

The Cartesian plane uses two perpendicular axes. The horizontal axis is labelled x, the vertical axis y, and they intersect at the origin (0, 0). Every point is written as an ordered pair (x, y), where x gives the horizontal displacement from the origin and y gives the vertical displacement. Points in the first quadrant have both coordinates positive. Points in the second quadrant have a negative x and positive y. The third quadrant has both negative. The fourth has positive x and negative y. Precision with signs matters: misreading a negative coordinate is one of the most common errors in this topic.

Gradient: measuring steepness

The gradient of a line measures how steeply it rises or falls. It is defined as the change in y divided by the change in x between any two points on the line:

gradient = (y₂ - y₁) / (x₂ - x₁)

A positive gradient means the line slopes upward from left to right. A negative gradient means it slopes downward. A gradient of zero produces a horizontal line. An undefined gradient (division by zero, because x₂ = x₁) produces a vertical line.

Worked example 1: finding a gradient

Problem: Find the gradient of the line passing through A(2, 3) and B(8, 15).

1. Identify the coordinates: x₁ = 2, y₁ = 3, x₂ = 8, y₂ = 15
2. Apply the formula: gradient = (15 - 3) / (8 - 2) = 12 / 6 = 2
3. Interpret: the line rises 2 units for every 1 unit it moves to the right

The order of the points does not matter, provided you subtract consistently. Using B first: (3 - 15) / (2 - 8) = (-12) / (-6) = 2. The result is identical. Mixing the order, taking y₂ - y₁ in the numerator but x₁ - x₂ in the denominator, produces a sign error. This is the single most frequent mistake candidates make with gradients.

Equations of straight lines

The equation y = mx + c describes every straight line that is not vertical. Here m is the gradient and c is the y-intercept, the value of y where the line crosses the y-axis (i.e., where x = 0). This form allows you to read off two critical pieces of information at a glance.

Worked example 2: equation from gradient and a point

Problem: A line has gradient 3 and passes through (4, 7). Find its equation.

1. Start with y = mx + c. Substitute m = 3: y = 3x + c
2. The point (4, 7) lies on the line, so substitute x = 4, y = 7: 7 = 3(4) + c
3. Solve: 7 = 12 + c, so c = -5
4. The equation is y = 3x - 5

Check by substituting the original point: y = 3(4) - 5 = 12 - 5 = 7. Correct.

Worked example 3: equation from two points

Problem: Find the equation of the line through P(-1, 5) and Q(3, -3).

1. Find the gradient: m = (-3 - 5) / (3 - (-1)) = -8 / 4 = -2
2. Substitute m = -2 and point P(-1, 5) into y = mx + c: 5 = -2(-1) + c
3. Solve: 5 = 2 + c, so c = 3
4. The equation is y = -2x + 3

Verification with Q: y = -2(3) + 3 = -6 + 3 = -3. Both points satisfy the equation, confirming the answer.

Drawing linear graphs

When a question asks you to draw a straight line, the table-of-values method is the most reliable approach:

1. Choose at least three x-values within the range given (three, not two, because a third point catches plotting errors)
2. Substitute each into the equation to find the corresponding y-value
3. Plot the points on the grid
4. Draw a straight line through them using a ruler

If the equation is not in y = mx + c form, rearrange it first. For example, 2x + 3y = 12 becomes y = (12 - 2x) / 3 = 4 - (2/3)x. Finding the x-intercept (set y = 0) and y-intercept (set x = 0) gives two reliable points: (6, 0) and (0, 4). A third point, say x = 3 giving y = 2, confirms the line.

Parallel and perpendicular lines

Two geometric relationships link gradients directly to the structure of lines on the Cartesian plane.

Parallel lines

Parallel lines never meet. They have the same gradient but different y-intercepts. If line L₁ has equation y = 2x + 5 and line L₂ has equation y = 2x - 1, both have gradient 2, so they are parallel. The converse also holds: if two distinct lines share a gradient, they are parallel.

Perpendicular lines (Extended)

Perpendicular lines meet at right angles. Their gradients are negative reciprocals of each other. The algebraic condition is:

m₁ x m₂ = -1

If one line has gradient 3, any line perpendicular to it has gradient -1/3. If one line has gradient -2/5, its perpendicular partner has gradient 5/2. The product of these two gradients always equals -1.

Worked example 4: finding a perpendicular equation

Problem: Line L has equation y = (2/3)x + 4. Find the equation of the line perpendicular to L that passes through (6, 1).

1. The gradient of L is 2/3
2. The perpendicular gradient satisfies m x (2/3) = -1, so m = -3/2
3. Substitute m = -3/2 and point (6, 1) into y = mx + c: 1 = (-3/2)(6) + c
4. Solve: 1 = -9 + c, so c = 10
5. The equation is y = (-3/2)x + 10

The logic flows in one direction: extract the gradient, invert and negate it, then use the point to find c. Candidates who try to shortcut this sequence often confuse the sign or forget to invert.

Length of a line segment

The distance between two points A(x₁, y₁) and B(x₂, y₂) is calculated using the distance formula, which is a direct application of Pythagoras' theorem:

d = sqrt((x₂ - x₁)² + (y₂ - y₁)²)

The horizontal distance |x₂ - x₁| and vertical distance |y₂ - y₁| form the two shorter sides of a right-angled triangle. The line segment AB is the hypotenuse.

Worked example 5: distance between two points

Problem: Find the distance between C(-3, 2) and D(5, -4).

1. Horizontal difference: 5 - (-3) = 8
2. Vertical difference: -4 - 2 = -6
3. Square both: 64 and 36
4. Sum: 64 + 36 = 100
5. Square root: sqrt(100) = 10

The distance is 10 units. Because each difference is squared, negative signs disappear automatically. You cannot get a negative distance, so if your answer is negative, check for a missing square root or an arithmetic error inside the radical.

Midpoint of a line segment

The midpoint of a line segment is the point exactly halfway between its two endpoints. The formula averages the coordinates:

Midpoint = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

This is intuitive: the midpoint of 3 and 9 on a number line is 6, which is (3 + 9) / 2. The same logic applies independently to both the x and y coordinates.

Worked example 6: finding a midpoint

Problem: Find the midpoint of E(2, 7) and F(10, -1).

1. Average the x-coordinates: (2 + 10) / 2 = 6
2. Average the y-coordinates: (7 + (-1)) / 2 = 3
3. The midpoint is (6, 3)

Worked example 7: finding an endpoint given the midpoint

Problem: The midpoint of G(1, 4) and H is (5, 2). Find H.

1. Let H = (a, b)
2. From the midpoint formula: (1 + a) / 2 = 5 and (4 + b) / 2 = 2
3. Solve for a: 1 + a = 10, so a = 9
4. Solve for b: 4 + b = 4, so b = 0
5. H = (9, 0)

This reverse application appears regularly on IGCSE papers. The method is identical to the forward calculation: set up the equation, solve. Candidates who memorise the midpoint formula without understanding it as an average struggle with this type of question.

Connecting the concepts

Exam questions rarely test a single skill in isolation. A typical extended question might give you two points, ask for the gradient, then the equation, then the perpendicular through a third point, and finally the distance between two intersections. The logical chain is always the same:

1. Extract or calculate a gradient
2. Use the gradient and a known point to build an equation
3. Apply a relationship (parallel, perpendicular, intersection) to derive the next piece
4. Use distance or midpoint formulas to answer a geometric question about the result

Each step feeds directly into the next. An error in step 1 propagates through every subsequent calculation, which is why the verification habit matters so much in this topic.

Common mistakes and how to avoid them

Self-check questions

1. Find the gradient of the line through (3, -2) and (7, 10). State whether the line slopes upward or downward from left to right.
2. A line has gradient -4 and passes through (2, 5). Write its equation in the form y = mx + c.
3. Find the equation of the line through (-1, 3) and (5, -9).
4. The line y = 5x - 2 is parallel to another line that passes through (0, 7). Write the equation of the parallel line.
5. A line has equation y = (1/4)x + 6. Find the gradient of a line perpendicular to it.
6. Find the equation of the line perpendicular to y = -2x + 1 that passes through (4, 3).
7. Calculate the distance between (-2, 5) and (4, -3). Give your answer as an exact value.
8. Find the midpoint of (7, -1) and (-3, 9).
9. The midpoint of A(2, 6) and B is (5, 1). Find the coordinates of B.
10. Two lines have equations y = 3x + 1 and y = -x + 9. Find the point where they intersect, then calculate the distance from this intersection to the origin.

Each question targets a specific skill from this topic. Work through them with full written solutions rather than mental arithmetic. The discipline of writing every step builds the accuracy that separates strong IGCSE candidates from those who understand the concepts but lose marks to avoidable slips.