Pearson Edexcel International GCSE Mathematics A 4MA1: Vectors and transformation geometry – Study Guide

This study guide teaches topic 5, Vectors and transformation geometry, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), first assessed in June 2018 with papers available in January and June. It covers section 5.1 (Vectors) and section 5.2 (Transformation geometry). Section 5.1 is Higher tier only. Section 5.2 is Foundation tier content, and the Higher tier assumes it (“See Foundation Tier”), so every candidate needs it.

Each tier sits two papers (1F and 2F, or 1H and 2H). Each paper lasts 2 hours, carries 100 marks and is worth 50% of the qualification. A calculator may be used on every paper, but most of this topic is exact counting on a grid, so show each step. None of the results below is on the formulae sheet.

See also the revision notes, the practice questions, the Edexcel IGCSE Mathematics hub, the printable 4MA1 checklist and the free 10-minute diagnostics.

What this topic covers

Section	What you must be able to do	Tier
5.1 A–B	Know that a vector has magnitude and direction; use the notations →OA, a and column vectors	Higher tier only
5.1 C–D	Multiply a vector by a scalar; add and subtract vectors	Higher tier only
5.1 E	Calculate the modulus (magnitude) of a vector	Higher tier only
5.1 F	Find the resultant of two or more vectors	Higher tier only
5.1 G	Use vector methods for simple geometrical proofs	Higher tier only
5.2 A–C	Rotate a shape about a centre through a given angle; anticlockwise is positive, clockwise is negative	Both tiers
5.2 D–E	Reflect a shape in a mirror line; construct a mirror line from an object and its image	Both tiers
5.2 F–H	Translate a shape; use column vectors to describe translations	Both tiers
5.2 I	Know that rotations, reflections and translations preserve length and angle, so the image is congruent	Both tiers
5.2 J–L	Enlarge a shape by a positive scale factor (including fractions), with or without a centre	Both tiers
5.2 M	Identify and give complete descriptions of transformations	Both tiers

Notation on this page. In the exam a vector from A to B is printed with an arrow over AB; here it is written →AB. A single-letter vector is printed in bold, a (underline it when you write by hand: a̲). The magnitude of a is |a|. A column vector is printed here on one line as (3, −2): the top number (across) comes first, the bottom number (up) second. In the exam it is written vertically.

5.1 Vectors (Higher tier only)

Magnitude and direction

A vector has both magnitude (size) and direction. A scalar, such as 5 or −2, has size only. Equal vectors have the same magnitude and direction, wherever they start.

A column vector (x, y) means “move x across and y up”. Negative x is left and negative y is down. So →AB = (4, −3) means from A go 4 right and 3 down to reach B. Reversing the direction changes every sign: →BA = −→AB = (−4, 3).

Scalar multiples, addition and subtraction

• Multiply by a scalar: 3(2, −5) = (6, −15). The result is parallel to the original and three times as long. A negative scalar reverses the direction.
• Add or subtract component by component: (2, −5) + (−1, 4) = (1, −1).
• Adding vectors means following one and then the other (“nose to tail”). The single vector from the start to the end is the resultant.

Two vectors are parallel when one is a scalar multiple of the other. For example, 6a − 9b = 3(2a − 3b), so 6a − 9b is parallel to 2a − 3b and three times as long.

Modulus

The modulus of (x, y) comes from Pythagoras’ theorem:

|(x, y)| = √(x² + y²)

Worked example 1. Find the magnitude of (7, −24).

|(7, −24)| = √(7² + (−24)²) = √(49 + 576) = √625 = 25

Square the negative component in full: (−24)² = +576.

Worked example 2. a = (4, −1) and b = (−2, 3). Find 2a − 3b and its magnitude.

2a  = (8, −2)
3b  = (−6, 9)
2a − 3b = (8 − (−6), −2 − 9) = (14, −11)
|2a − 3b| = √(14² + (−11)²) = √(196 + 121) = √317 = 17.8 (3 s.f.)

Resultants in a diagram

To find a vector between two points, walk along a route of known vectors. Add a vector when you travel along its arrow. Subtract it when you travel against it.

The specification’s own illustration uses →OA = 3a, →AB = 2b and →BC = c, so →OC = 3a + 2b + c and →CA = −c − 2b.

Worked example 3. In a quadrilateral PQRS, →PQ = a + 2b, →QR = 3a − b and →RS = −2a. Find →PS and →SQ.

PS = PQ + QR + RS = (a + 2b) + (3a − b) + (−2a) = 2a + b
SQ = SR + RQ = 2a − (3a − b) = −a + b

Checking by another route, →SP + →PQ, gives the same answer.

Vector proof

Section 5.1 G asks you to “apply vector methods for simple geometrical proofs”. The usual targets are:

• Parallel lines. Show that one vector is a scalar multiple of the other.
• Points on a straight line (collinear). Show that →XY and →YZ (or →XY and →XZ) are multiples of each other and share the point Y (or X).
• Ratios along a line. If P divides AB in the ratio m : n, then →AP = (m/(m + n))→AB.

Worked example 4. OAB is a triangle with →OA = a and →OB = b. N lies on AB with AN : NB = 1 : 2. The point D satisfies →OD = 2a + b. Prove that O, N and D lie on a straight line, and find ON : ND.

AB = AO + OB = −a + b = b − a
AN = (1/3)AB = (1/3)(b − a)
ON = OA + AN = a + (1/3)b − (1/3)a = (2/3)a + (1/3)b = (1/3)(2a + b)
OD = 2a + b = 3 × ON

→OD is a multiple of →ON, so the lines ON and OD are parallel. They share the point O, so O, N and D are collinear. Also →ND = →OD − →ON = (2/3)(2a + b) = 2→ON, so ON : ND = 1 : 2.

A proof must end with a sentence that names both facts: the scalar multiple and the common point. “Parallel” alone does not prove three points are on one line.

5.2 Transformation geometry (both tiers)

A transformation maps an object onto an image. Label the image clearly, for example A′ for the image of A.

Rotation

A rotation is specified by a centre and an angle. An anticlockwise rotation is a positive angle; a clockwise rotation is negative. So “rotation through −90°” means 90° clockwise. A rotation of 180° needs no direction.

On a grid, use tracing paper pinned at the centre. With coordinates, subtract the centre, turn the vector, then add the centre back.

Rotation about the origin	Point (x, y) maps to
+90° (anticlockwise)	(−y, x)
−90° (clockwise)	(y, −x)
180°	(−x, −y)

Worked example 5. Triangle T has vertices (3, 2), (5, 2) and (3, 5). Rotate T through +90° about (2, 1).

Subtract the centre:   (1, 1), (3, 1), (1, 4)
Turn +90°: (x, y) → (−y, x)   gives (−1, 1), (−1, 3), (−4, 1)
Add the centre back:   (1, 2), (1, 4), (−2, 2)

The image has vertices (1, 2), (1, 4) and (−2, 2). Check one length: the side from (3, 2) to (5, 2) has length 2, and so does its image from (1, 2) to (1, 4).

Reflection

A reflection is specified by a mirror line. The specification lists lines such as x = 1, y = 2, y = x and y − x = 0. Note that y − x = 0 is the same line as y = x.

• In x = a: (x, y) maps to (2a − x, y).
• In y = b: (x, y) maps to (x, 2b − y).
• In y = x: (x, y) maps to (y, x).

For example, reflecting (3, −1) gives (−1, 3) in y = x, (−1, −1) in x = 1 and (3, 5) in y = 2. Each image point is the same perpendicular distance from the line as its object, on the other side. A point on the mirror line stays where it is.

To construct a mirror line from an object and its image, join a point to its image and draw the perpendicular bisector of that segment. Check it with a second pair of points.

Translation

A translation is specified by a distance and a direction, written as a column vector. Translating by (−5, 3) moves every point 5 left and 3 up. To describe a translation, subtract object coordinates from image coordinates: image − object.

What each transformation keeps

Rotations, reflections and translations preserve length and angle, so the image is congruent to the object. Enlargements preserve angles but not lengths, so the image is similar to the object, not congruent (unless the scale factor is 1).

Enlargement

An enlargement is specified by a centre and a scale factor. In 4MA1 the scale factor is positive, and it may be a fraction. A scale factor between 0 and 1 makes the image smaller and closer to the centre. The specification also expects enlargement without a given centre: then only the size changes, and you may draw the image anywhere.

With a centre C, measure each vertex from C and multiply by the scale factor:

image = C + k × (P − C)

Worked example 6. Enlarge the triangle with vertices (4, 2), (8, 2) and (8, 6) by scale factor 1/2, centre (2, 0).

P − C:            (2, 2), (6, 2), (6, 6)
× 1/2:            (1, 1), (3, 1), (3, 3)
add C back:       (3, 1), (5, 1), (5, 3)

The image is (3, 1), (5, 1), (5, 3). Each side is half as long and each angle is unchanged.

Complete descriptions

Section 5.2 M asks for complete descriptions. Each mark scheme point is a separate fact, so give all of them and give only one transformation.

Transformation	A complete description states
Rotation	“Rotation”, the angle (with direction or sign), the centre
Reflection	“Reflection”, the equation of the mirror line
Translation	“Translation”, the column vector
Enlargement	“Enlargement”, the scale factor, the centre

Worked example 7. Shape P has vertices (2, 1), (3, 1) and (2, 3). Shape Q has vertices (4, 1), (7, 1) and (4, 7). Describe fully the single transformation that maps P onto Q.

The side from (2, 1) to (3, 1) (length 1) maps to the side from (4, 1) to (7, 1) (length 3): same way round, three times larger, so scale factor 3. On a grid, draw lines through each vertex and its image; they meet at the centre. By algebra, Q = C + 3(P − C) gives C = (3P − Q)/2. Using P = (2, 1) and Q = (4, 1) gives C = ((6 − 4)/2, (3 − 1)/2) = (1, 1). Check with (3, 1) mapping to (7, 1): (1, 1) + 3 × (2, 0) = (7, 1). ✓

Answer: enlargement, scale factor 3, centre (1, 1).

Common errors

• Writing →AB as A − B instead of B − A. The vector from A to B is “end minus start”.
• Squaring −11 as −121 inside a modulus. Every square is positive.
• Declaring three points collinear after showing only that two vectors are parallel, with no common point named.
• Using “a rotation of 90°” with no direction. The sign or the word clockwise/anticlockwise is part of the answer.
• Rotating about the origin when the centre given is a different point.
• Reflecting in y = x by changing signs instead of swapping coordinates.
• Describing a translation in words (“5 left, 3 up”) instead of a column vector, or writing the two components the wrong way round.
• Measuring an enlargement from the origin instead of the stated centre.
• Giving two transformations (for example “reflection then translation”) when the question says “single transformation”. This scores nothing.

Next steps

Recap with the revision notes, then try the practice questions.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification Issue 2, November 2017, Pearson Education Limited. Topic 5, Vectors and transformation geometry: 5.1 Vectors (Higher tier only) and 5.2 Transformation geometry.