Rectangular Cartesian coordinates for Edexcel IGCSE Further Pure Mathematics: Methods and Worked Examples

Rectangular Cartesian coordinates are the foundation of coordinate geometry, and in the edexcel igcse further pure mathematics exam they underpin questions on straight lines, calculus and curve sketching

If you know the coordinates of two points, you can find the distance between them, the midpoint of the segment joining them, the gradient of the line through them, and the equation of that line. These are core skills that the exam tests directly and that also appear as intermediate steps in calculus, trigonometry and vector questions. Coordinate geometry is the bridge between algebra and geometry: everything you prove algebraically here has a geometric interpretation, and everything you draw can be verified algebraically.

These edexcel igcse further pure mathematics rectangular cartesian coordinates revision notes cover the full section with worked examples and no-nonsense exam strategy. The approach is direct: learn the formula, see it used, then practise until it becomes reflexive.

Distance between two points

The distance between points A(x₁, y₁) and B(x₂, y₂) is:

d = \u221a((x₂ - x₁)² + (y₂ - y₁)²)

Worked example: Find the distance between A(3, -1) and B(-2, 7).

d = \u221a((-2 - 3)² + (7 - (-1))²) = \u221a(25 + 64) = \u221a89.

Leave in surd form unless told otherwise. The exam often asks for exact answers. Writing \u221a89 is exact; writing 9.43 is approximate and loses the accuracy mark. Only round to a given number of significant figures or decimal places when the question explicitly instructs you to.

Midpoint and section formula

The midpoint of the segment joining A(x₁, y₁) and B(x₂, y₂) is:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Worked example: Find the midpoint of A(4, -3) and B(10, 5).

M = ((4 + 10)/2, (-3 + 5)/2) = (7, 1).

Dividing a line in a given ratio

If P divides the line segment AB in the ratio m:n, then:

P = ((nx₁ + mx₂)/(m + n), (ny₁ + my₂)/(m + n))

Worked example: Find the coordinates of the point that divides the segment from A(1, 2) to B(7, 14) in the ratio 2:1.

P = ((1(1) + 2(7))/(2 + 1), (1(2) + 2(14))/(2 + 1)) = ((1 + 14)/3, (2 + 28)/3) = (5, 10).

Gradient of a straight line

The gradient of the line joining A(x₁, y₁) and B(x₂, y₂) is:

m = (y₂ - y₁) / (x₂ - x₁)

Worked example: Find the gradient of the line through (2, 5) and (6, -3).

m = (-3 - 5) / (6 - 2) = -8/4 = -2.

Parallel and perpendicular lines

Condition	Gradient relationship
Parallel lines	m₁ = m₂ (equal gradients)
Perpendicular lines	m₁ x m₂ = -1 (product of gradients is -1)

Worked example: Line L₁ has gradient 3/4. Find the gradient of a line perpendicular to L₁.

m₂ = -1 / (3/4) = -4/3.

Equations of straight lines

The specification requires you to use two forms of the equation of a straight line:

y = mx + c: gradient-intercept form, where m is the gradient and c is the y-intercept.
y - y₁ = m(x - x₁): point-gradient form, where (x₁, y₁) is a known point on the line.

You also need to interpret ax + by = c as a straight line equation.

Worked example: finding the equation of a line

Find the equation of the line through (3, -2) with gradient 5, giving your answer in the form ax + by + c = 0.

y - (-2) = 5(x - 3)

y + 2 = 5x - 15

5x - y - 17 = 0.

Worked example: perpendicular bisector

Find the equation of the perpendicular bisector of AB, where A = (1, 3) and B = (5, 11).

Midpoint M = ((1 + 5)/2, (3 + 11)/2) = (3, 7).

Gradient of AB = (11 - 3)/(5 - 1) = 8/4 = 2.

Gradient of perpendicular bisector = -1/2.

Equation: y - 7 = (-1/2)(x - 3), so y = -x/2 + 3/2 + 7 = -x/2 + 17/2.

In standard form: x + 2y - 17 = 0.

Worked example: intersection of two lines

Find the point of intersection of y = 2x + 1 and 3x + y = 11.

Substitute y = 2x + 1 into 3x + y = 11: 3x + 2x + 1 = 11, so 5x = 10, x = 2. Then y = 2(2) + 1 = 5.

The lines intersect at (2, 5).

Read the required form carefully. The exam may ask for the equation in the form y = mx + c, or in the form ax + by + c = 0, or even in the form ax + by = c. Presenting the correct answer in the wrong form can cost you the final mark. Always check the format specified in the question before writing your final answer.

Worked examples: multi-step coordinate geometry problems

The exam frequently sets questions that combine several coordinate geometry skills in a single problem. Practising these builds the fluency needed for the edexcel igcse further pure mathematics exam.

Worked example: triangle and perpendicular from a vertex

Triangle ABC has vertices A(1, 2), B(7, 4) and C(3, 8). Find the equation of the altitude from C to AB, and find the foot of the perpendicular.

Gradient of AB = (4 - 2)/(7 - 1) = 2/6 = 1/3.

The altitude from C is perpendicular to AB, so its gradient = -3.

Equation: y - 8 = -3(x - 3), so y = -3x + 9 + 8 = -3x + 17.

To find the foot of the perpendicular, solve simultaneously with the line AB. AB has equation y - 2 = (1/3)(x - 1), so y = x/3 + 5/3.

Set -3x + 17 = x/3 + 5/3. Multiply through by 3: -9x + 51 = x + 5, so -10x = -46, giving x = 23/5.

y = (23/5)/3 + 5/3 = 23/15 + 25/15 = 48/15 = 16/5.

The foot of the perpendicular is (23/5, 16/5).

Worked example: collinearity test using gradients

Show that the points P(-1, 5), Q(3, 3) and R(11, -1) are collinear.

Gradient PQ = (3 - 5)/(3 - (-1)) = -2/4 = -1/2.

Gradient QR = (-1 - 3)/(11 - 3) = -4/8 = -1/2.

Since PQ and QR have the same gradient and share point Q, the three points are collinear.

Combining coordinate geometry with other topics

In the rectangular cartesian coordinates edexcel igcse specification, these skills rarely appear in isolation. The exam commonly pairs coordinate geometry with:

Calculus: find the equation of the tangent or normal to a curve at a given point (uses gradient and point-gradient form).
Circles: show that a point lies on a circle using the distance formula, or find the equation of a tangent to a circle.
Vectors: verify collinearity using gradients, or find the position of a point dividing a segment.

Practice questions

These igcse 4PM1 rectangular cartesian coordinates problems test the key techniques from this section. Attempt each one before reading the solution.

Find the distance between (-3, 4) and (5, -2), giving your answer in simplified surd form.
The line L passes through (2, -1) and is perpendicular to the line 4x - 3y + 7 = 0. Find the equation of L in the form ax + by + c = 0.
A(1, 5) and B(9, 1). Find the coordinates of the point P on AB such that AP:PB = 3:1.
Find the area of the triangle with vertices A(0, 0), B(6, 0) and C(4, 5).
The perpendicular bisector of the segment joining (2, 8) and (6, 2) passes through the point (k, 0). Find k.

Solutions

Question 1: d = \u221a((5-(-3))² + (-2-4)²) = \u221a(64 + 36) = \u221a100 = 10.

Question 2: Rearrange 4x - 3y + 7 = 0 to y = (4/3)x + 7/3. Gradient = 4/3. Perpendicular gradient = -3/4. Equation: y - (-1) = (-3/4)(x - 2), so y + 1 = -3x/4 + 3/2, giving 3x + 4y - 2 = 0.

Question 3: P = ((1(1) + 3(9))/4, (1(5) + 3(1))/4) = (28/4, 8/4) = (7, 2).

Question 4: Using the base-height formula: base AB = 6, height = y-coordinate of C = 5. Area = (1/2)(6)(5) = 15 square units.

Question 5: Midpoint = (4, 5). Gradient of segment = (2 - 8)/(6 - 2) = -6/4 = -3/2. Perpendicular gradient = 2/3. Equation of bisector: y - 5 = (2/3)(x - 4). Setting y = 0: -5 = (2/3)(x - 4), so x - 4 = -15/2, giving x = 4 - 15/2 = -7/2. Therefore k = -7/2.

The edexcel igcse further pure mathematics notes for coordinate geometry should be among the most well-practised in your revision folder. The formulae are straightforward, but the edexcel igcse further pure mathematics explained skill lies in applying them accurately and in the correct form under exam conditions. For more edexcel igcse further pure mathematics practice questions on coordinates and every other specification topic, use edexcel igcse further pure mathematics revision notes on the Green Bridge CBT platform to track your progress and eliminate recurring errors.