Analytical Geometry — Grade 10 Maths (CAPS)

The three core formulas — distance, midpoint, gradient — and the ways Grade 10 exams use them: parallel and perpendicular lines, collinear points, and finding an unknown coordinate, each worked in full with a worksheet.

Analytical (coordinate) geometry works out geometric facts using algebra, by placing points on the Cartesian plane. Almost every question uses one of three formulas, applied to two points (x₁ ; y₁) and (x₂ ; y₂). This guide covers the formulas and every question type built on them.

1The three formulas

Distance d = √((x₂ − x₁)² + (y₂ − y₁)²)

Midpoint M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 ) • Gradient m = (y₂ − y₁)/(x₂ − x₁) (x₂ ≠ x₁)

Distance uses the horizontal gap Δx and vertical gap Δy (Pythagoras); gradient is Δy ÷ Δx.

2Type 1: Distance, midpoint and gradient of two points

Worked ExampleExample 1: A(1 ; 2) and B(4 ; 6)

1
Distance: d = √((4 − 1)² + (6 − 2)²) = √(9 + 16) = √25 = 5.
Pythagoras on the horizontal gap (3) and vertical gap (4).
2
Midpoint: (1 + 4)/2 , (2 + 6)/2 = (2,5 ; 4).
Average each coordinate.
3
Gradient: m = (6 − 2)/(4 − 1) = 4/3.
Rise over run — keep the points in the same order.

3Type 2: Parallel lines (equal gradients)

Two lines are parallel when their gradients are equal.

Worked ExampleExample 2: is the line through A(0 ; 1), B(2 ; 5) parallel to the one through C(1 ; 0), D(3 ; 4)?

1
Gradient of AB: (5 − 1)/(2 − 0) = 4/2 = 2.
Find each gradient.
2
Gradient of CD: (4 − 0)/(3 − 1) = 4/2 = 2.
3
Equal gradients (both 2) → the lines are parallel.

4Type 3: Perpendicular lines (gradients multiply to −1)

Two lines are perpendicular when their gradients multiply to −1 (each is the negative reciprocal of the other). One exception: a vertical line has no gradient, so the product rule cannot be used for it: a line perpendicular to a vertical line is horizontal (gradient 0), and the other way round. For two non-vertical lines, perpendicular means m₁ × m₂ = −1.

Worked ExampleExample 3: a line has gradient 3. What gradient is perpendicular to it?

1
Perpendicular gradient × 3 = −1.
Product of perpendicular gradients is −1.
2
So the perpendicular gradient = −1/3.
The negative reciprocal of 3.

5Type 4: Collinear points (on the same straight line)

Worked ExampleExample 4: are A(1 ; 2), B(3 ; 6), C(5 ; 10) collinear?

1
Gradient AB: (6 − 2)/(3 − 1) = 4/2 = 2.
Points are collinear if the gradients between them are equal.
2
Gradient BC: (10 − 6)/(5 − 3) = 4/2 = 2.
3
Equal gradients → the three points are collinear.
They lie on one straight line.

6Type 5: Find an unknown coordinate

Worked ExampleExample 5: the midpoint of A(2 ; 3) and B(x ; 9) is (5 ; 6). Find x

1
The midpoint x-value: (2 + x)/2 = 5.
Use the midpoint formula on the x-coordinates.
2
2 + x = 10 → x = 8.
Solve the equation.

Worked ExampleExample 6: the distance from (0 ; 0) to (x ; 4) is 5. Find x (x > 0)

1
√((x − 0)² + (4 − 0)²) = 5 → x² + 16 = 25.
Square both sides to remove the root.
2
x² = 9 → x = 3.
Take the positive root (given x > 0).

💡

Parallel → equal gradients; perpendicular → gradients multiply to −1. These two facts unlock most 'prove it' questions about shapes (e.g. showing a quadrilateral is a rectangle).

⚠️

Keep the points in the same order on the top and bottom of the gradient fraction. Swapping only x or only y flips the sign and gives the wrong gradient.

Analytical geometry expands to lines, circles and tangents in Grades 11–12, all built on these three formulas. Master every type below.

7Proving a shape from its coordinates

The ATP asks you to represent quadrilaterals on the Cartesian plane and then prove what they are. You already have every tool: distance decides whether sides are equal, gradient decides whether they are parallel or perpendicular, and the midpoint decides whether diagonals bisect each other.

Two sides equal? → distance formula on both.
Two sides parallel? → gradients equal.
Perpendicular? → product of the gradients is −1.
Diagonals bisect each other? → both diagonals have the same midpoint.

Worked ExampleA(1 ; 2), B(5 ; 4), C(7 ; 8) and D(3 ; 6). Prove that ABCD is a parallelogram.

1
m(AB) = (4 − 2)/(5 − 1) = 2/4 = ½
Gradient of one side.
2
m(DC) = (8 − 6)/(7 − 3) = 2/4 = ½
The opposite side.
3
AB ∥ DC
Equal gradients.
4
AB = √((5 − 1)² + (4 − 2)²) = √20 and DC = √((7 − 3)² + (8 − 6)²) = √20
Distance formula on the same pair.
5
∴ ABCD is a parallelogram
One pair of opposite sides equal AND parallel.

💡

Prove the cheapest condition. Diagonals that bisect each other need only two midpoints. One pair of opposite sides that is both equal and parallel needs two gradients and two distances. Four equal sides also proves it — a quadrilateral with all four sides equal is a rhombus, and every rhombus is a parallelogram — but that takes four distance calculations.

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Deciding the shape by looking at the plotted points. A sketch is for orientation only — the marks are for the calculation. And check the vertices are given in order round the shape: A, B, C, D means AB and DC are the opposite pair, not AB and CD reversed.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Distance between (1 ; 1) and (4 ; 5). (distance)
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√(9 + 16) = √25 = 5
2
Distance between (−2 ; 1) and (1 ; 5). (distance)
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√(9 + 16) = 5
3
Midpoint of (2 ; 4) and (6 ; 10). (midpoint)
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(2 + 6)/2,(4 + 10)/2 = (4 ; 7)
4
Gradient of the line through (1 ; 2) and (5 ; 10). (gradient)
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(10 − 2)/(5 − 1) = 2
5
Are gradients 3 and 3 parallel or perpendicular? (parallel)
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Equal → parallel
6
What gradient is perpendicular to gradient 2? (perpendicular)
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−1/2 (negative reciprocal)
7
What gradient is perpendicular to gradient −3/4? (perpendicular)
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4/3
8
Are (0 ; 0), (2 ; 4), (3 ; 6) collinear? (collinear)
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Gradients 2 and 2 → Yes
9
Midpoint of (1 ; 2) and (x ; 8) is (4 ; 5). Find x. (unknown)
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(1 + x)/2 = 4 → x = 7
10
Distance from (0 ; 0) to (x ; 12) is 13 (x > 0). Find x. (unknown)
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x² + 144 = 169 → x = 5
11
P(−1 ; 1), Q(3 ; 3), R(5 ; 7) and S(1 ; 5) are the vertices of a quadrilateral. Prove that PQRS is a parallelogram.
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m(PQ) = 2/4 = ½ ✓; m(SR) = 2/4 = ½ ✓, so PQ ∥ SR; PQ = √20 and SR = √20 ✓; ∴ parallelogram (one pair opp sides equal and parallel) ✓
12
A(0 ; 0), B(4 ; 2) and C(2 ; 6). Show that AB ⊥ BC.
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m(AB) = 2/4 = ½ ✓; m(BC) = (6 − 2)/(2 − 4) = −2 ✓; ½ × (−2) = −1, so AB ⊥ BC ✓
13
The diagonals of quadrilateral KLMN are KM and LN. K(1 ; 1), L(6 ; 2), M(7 ; 6), N(2 ; 5). Show that the diagonals bisect each other.
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Midpoint KM = (4 ; 3,5) ✓; midpoint LN = (4 ; 3,5) ✓; the midpoints are the same, so the diagonals bisect each other ✓