Pearson Edexcel International GCSE Mathematics A 4MA1: Vectors and transformation geometry – Revision Notes

These revision notes cover 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 in January and June. They cover section 5.1 Vectors, which is Higher tier only, and section 5.2 Transformation geometry, which is Foundation content assumed at Higher tier, so both tiers need it. For full explanations and worked examples, read the study guide first.

Other links: practice questions with worked answers, Edexcel IGCSE Mathematics hub, printable 4MA1 checklist, free 10-minute diagnostics.

A calculator may be used on all four 4MA1 papers (1F, 2F, 1H, 2H). None of the rules below is printed on the formulae sheet, so learn them.

Notation: →AB is the vector from A to B (arrow over AB in the exam); a is a vector (underline it by hand); |a| is its magnitude; (x, y) written in a vector context is the column vector with x on top.

5.1 Vectors (Higher tier only)

Key definitions

• Vector: a quantity with magnitude and direction.
• Scalar: a quantity with magnitude only.
• Resultant: the single vector with the same effect as two or more vectors followed in turn.
• Modulus: the magnitude (length) of a vector, written |a|.
• Parallel vectors: one is a scalar multiple of the other.

Rules

Operation	Rule	Example
Scalar multiple	k(x, y) = (kx, ky)	−2(3, −1) = (−6, 2)
Addition	(a, b) + (c, d) = (a + c, b + d)	(1, 4) + (2, −6) = (3, −2)
Subtraction	(a, b) − (c, d) = (a − c, b − d)	(1, 4) − (2, −6) = (−1, 10)
Reverse	→BA = −→AB	→AB = (5, 2) gives →BA = (−5, −2)
Between points	→AB = →OB − →OA (“end minus start”)	A(1, 3), B(4, −1): →AB = (3, −4)
Modulus	|(x, y)| = √(x² + y²)	|(3, −4)| = 5
Ratio on a line	If AP : PB = m : n, →AP = (m/(m + n))→AB	AP : PB = 3 : 1 gives →AP = (3/4)→AB

Method: finding a vector in a diagram

1. Choose a route from the start point to the end point using only known vectors.
2. Add each vector you travel along; subtract each one you travel against.
3. Collect like terms (a terms, then b terms).
4. If the route passes a fraction of a side, first write that side as a vector, then take the fraction.

Small reminder: if →OA = a, →OB = b and M is the midpoint of AB, then →OM = a + ½(b − a) = ½a + ½b.

Method: vector proof

1. Write every vector you need in terms of the same base vectors (usually a and b).
2. For parallel: show →PQ = k→RS for a number k. Write the factor out, for example 4a − 6b = 2(2a − 3b).
3. For collinear P, Q, R: show →PQ = k→QR (or k→PR) and state that the two vectors share the point Q (or P).
4. Finish with a sentence: “→PQ is a multiple of →QR and Q is a common point, so P, Q and R lie on a straight line.”
5. If a ratio is asked, read it off from k: →PQ = 2→QR gives PQ : QR = 2 : 1.

Small worked reminder: in triangle OAB, →OA = a and →OB = b. M is on OA with OM : MA = 1 : 2, and N is on OB with ON : NB = 1 : 2.

OM = (1/3)a        ON = (1/3)b
MN = MO + ON = −(1/3)a + (1/3)b = (1/3)(b − a)
AB = b − a
MN = (1/3)AB

So MN is parallel to AB and one third of its length. No common point is needed here, because the question is about parallel lines, not collinear points.

Must-know distinctions

• →AB vs →BA: same length, opposite direction; every component changes sign.
• Parallel vs collinear: parallel needs a scalar multiple only; collinear also needs a shared point.
• Vector vs modulus: a vector has two components; its modulus is one positive number.
• Position vector vs displacement: →OA is the position vector of A; →AB is a displacement between two points.

5.2 Transformation geometry (both tiers)

What defines each transformation

Transformation	Specified by	Preserves	Image is
Rotation	Centre and angle (anticlockwise positive, clockwise negative)	Length and angle	Congruent
Reflection	Mirror line	Length and angle	Congruent
Translation	Column vector (distance and direction)	Length and angle	Congruent
Enlargement	Centre and positive scale factor (fractions allowed)	Angle only	Similar, not congruent unless the scale factor is 1

Coordinate rules

Transformation	(x, y) maps to
Rotation +90° about O	(−y, x)
Rotation −90° about O	(y, −x)
Rotation 180° about O	(−x, −y)
Reflection in y = x (same line as y − x = 0)	(y, x)
Reflection in x = a	(2a − x, y)
Reflection in y = b	(x, 2b − y)
Translation by (p, q)	(x + p, y + q)
Enlargement, scale factor k, centre (c, d)	(c + k(x − c), d + k(y − d))

For a rotation about a centre that is not O: subtract the centre, apply the rule, add the centre back.

Small reminder: rotating (2, 5) through −90° about O gives (5, −2). Swap, then change the sign of the new second coordinate.

Method: describe fully a single transformation

1. Same size and same way round, just moved: translation. Give the column vector (image − object).
2. Same size, flipped (a mirror image): reflection. Give the equation of the mirror line. Find it as the perpendicular bisector of a point and its image.
3. Same size, turned: rotation. Give the angle with a sign or direction, and the centre. Find the centre with tracing paper, or where the perpendicular bisectors of two object–image segments meet.
4. Different size: enlargement. Scale factor = image length ÷ object length. Find the centre by drawing lines through each vertex and its image; they meet at the centre.
5. Write one transformation only, using the word itself (“rotation”, not “turn”).

Small worked reminder: a triangle with a side of 6 cm maps to a triangle whose matching side is 2 cm, the same way round. The scale factor is 2 ÷ 6 = 1/3, so the answer starts “enlargement, scale factor 1/3”. The image is smaller, but it is still called an enlargement; 4MA1 uses positive scale factors only.

Must-know distinctions

• Clockwise vs anticlockwise: −90° is clockwise; +90° is anticlockwise.
• x = 2 vs y = 2: x = 2 is a vertical line; y = 2 is horizontal.
• Scale factor ½ vs 2: ½ makes the image smaller and nearer the centre.
• Congruent vs similar: rotation, reflection and translation give congruent images; enlargement gives similar images.

Quick self-test

1. Find |(−9, 12)|.
2. a = (3, 5) and b = (−1, 2). Find a − 2b.
3. →AB = (2, −7). Write down →BA.
4. |(k, 12)| = 13 and k > 0. Find k.
5. Is 6a − 9b parallel to 2a − 3b? Give a reason.
6. Reflect (5, −2) in the line y = x.
7. Reflect (5, −2) in the line x = 1.
8. Rotate (4, 1) through 180° about the origin.
9. Rotate (4, 1) through +90° about the origin.
10. Enlarge the point (6, 9) by scale factor 1/3, centre (0, 3).
11. Translate (−3, 4) by the vector (5, −6).
12. Rotate (3, 4) through −90° about (1, 1).

Answers

1. √(81 + 144) = √225 = 15
2. (3, 5) − (−2, 4) = (5, 1)
3. (−2, 7)
4. k² + 144 = 169, so k² = 25 and k = 5
5. Yes: 6a − 9b = 3(2a − 3b), a scalar multiple.
6. (−2, 5)
7. The point is 4 units right of x = 1, so the image is 4 units left: (−3, −2)
8. (−4, −1)
9. (−1, 4)
10. (0, 3) + (1/3)(6, 6) = (2, 5)
11. (2, −2)
12. Subtract centre: (2, 3). Turn −90°: (3, −2). Add centre: (4, −1)

Where marks are usually lost

• Answering “rotation, 90°” with no direction or sign, which usually loses the angle mark even when the centre is right.
• Giving the centre of rotation or enlargement as a vector, or leaving it out; the centre must be a coordinate point.
• Naming two transformations when the question asks for a single transformation; that answer usually scores no marks.
• Reflecting in x = 1 as though it were y = 1 (or the reverse). Sketch the line first.
• Finding a translation vector as object − image, which reverses both signs.
• Enlarging by a fractional scale factor from the origin when a different centre is given.
• In vector work, writing →AB as a − b when →OA = a and →OB = b; the correct vector is b − a.
• Leaving a modulus as 17.80449… when the question asks for 3 significant figures, or rounding too early in a longer calculation.
• In a “show that” or proof, simplifying correctly but never writing the concluding sentence about parallel lines or the common point.
• Treating a vector expression like 2a + b as a number and trying to “cancel” a with b.

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.