9 min readConcept sheet3-mark questions; a quadrilateral proof can be a 5-mark answer or part of a case-based question

Collinearity and the type of a quadrilateral

Use distances to test shapes. Three points are collinear if the two shorter distances add up to the longest; sides and diagonals name the quadrilateral.

Collinear points A, B, CAB + BC = AC, where AC is the longest of the three distances

Square4 equal sides and equal diagonals

Rhombus (not a square)4 equal sides but unequal diagonals

Rectangle and parallelogramOpposite sides equal. Equal diagonals: rectangle. Diagonals with the same mid-point: parallelogram

Which quadrilateral? Test with distances

Sides	Diagonals	Quadrilateral
All 4 equal	Equal	Square
All 4 equal	Not equal	Rhombus (not a square)
Opposite sides equal	Equal	Rectangle
Opposite sides equal	Not equal	Parallelogram (not a rectangle)
Any	Same mid-point (they bisect each other)	At least a parallelogram

Example 1: are the points collinear?

Problem

Check whether A(1, −1), B(4, 3) and C(7, 7) are collinear.

Working

Distance formula: d = √[(x₂ − x₁)² + (y₂ − y₁)²]
AB = √[(4 − 1)² + (3 + 1)²] = √(9 + 16) = 5
BC = √[(7 − 4)² + (7 − 3)²] = √(9 + 16) = 5
AC = √[(7 − 1)² + (7 + 1)²] = √(36 + 64) = 10
AB + BC = 5 + 5 = 10 = AC
So A, B and C lie on one straight line.

Answer

Yes, A, B and C are collinear.

Example 2: name the quadrilateral

Problem

Show that A(1, 2), B(4, 3), C(5, 6) and D(2, 5) are the vertices of a rhombus. Is it a square?

Working

Use the distance formula for each side.
AB = √[(4 − 1)² + (3 − 2)²] = √(9 + 1) = √10
BC = √[(5 − 4)² + (6 − 3)²] = √(1 + 9) = √10
CD = √[(2 − 5)² + (5 − 6)²] = √(9 + 1) = √10
DA = √[(1 − 2)² + (2 − 5)²] = √(1 + 9) = √10
All four sides are equal, so ABCD is a rhombus.
Diagonals: AC = √(4² + 4²) = √32 and BD = √[(2 − 4)² + (5 − 3)²] = √8
AC ≠ BD, so ABCD is not a square.

Answer

ABCD is a rhombus, but not a square.

Example 3: the fourth vertex of a parallelogram

Problem

A(1, 2), B(4, 3) and C(6, 6) are three vertices of parallelogram ABCD. Find D.

Working

The diagonals of a parallelogram bisect each other, so AC and BD have the same mid-point.
Mid-point of AC = ((1 + 6)/2, (2 + 6)/2) = (7/2, 4)
Let D be (x, y). Mid-point of BD = ((4 + x)/2, (3 + y)/2)
(4 + x)/2 = 7/2, so x = 3
(3 + y)/2 = 4, so y = 5

Answer

D is (3, 5).

Also asked: types of triangles

Isosceles: two sides equal. Equilateral: all three sides equal.
Right-angled: the square of the longest side equals the sum of the squares of the other two. This is the converse of Pythagoras' theorem, known from earlier classes: you use it here, you do not prove it.
To show a point is the centre of a circle through three points, show that it is the same distance from all three.
Find all the lengths you need before you decide. One or two lengths are not enough.

In real life

Before marking a kabaddi or kho-kho court, people check the rectangle by measuring both diagonals. Equal diagonals (with equal opposite sides) mean the corners are right angles.

Exam angle

2 marks: “A(2, 1), B(5, 2), C(6, 4) and D(x, y) form parallelogram ABCD, in that order. Find D.” The diagonals bisect each other. Mid-point of AC = (4, 5/2). So (5 + x)/2 = 4 and (2 + y)/2 = 5/2, giving D(3, 3).

3 marks: “Check whether the points (−2, 3), (1, 0) and (4, −3) are collinear.” Call them A, B and C. AB = √(9 + 9) = 3√2, BC = 3√2 and AC = √(36 + 36) = 6√2. AB + BC = AC, so the points are collinear.

5 marks: “Show that A(−1, 0), B(3, 1), C(2, 5) and D(−2, 4) are the vertices of a square.” AB = BC = CD = DA = √17 and both diagonals AC = BD = √34. Equal sides and equal diagonals, so ABCD is a square.

Remember it

Collinear: small + small = big. Square: sides same, diagonals same. Rhombus: sides same, diagonals differ.