Mathematics 0580Chapter 7 of 9

Transformations and Vectors

Cambridge IGCSE Mathematics 0580 Topic 7 revision chapter covering Transformations and Vectors for the 2025-2027 syllabus, version 3, with the Core and Extended routes separated throughout. Only C7.1 Transformations is active Core content; C7.2, C7.3 and C7.4 are Extended content only, so the vector half of the topic is studied by Extended candidates alone. The Core foundation teaches reflection of a shape in a vertical or horizontal line, rotation of a shape about the origin, a vertex or the midpoint of an edge through multiples of ninety degrees, enlargement of a shape from a centre by a positive or fractional scale factor, and translation of a shape by a column vector, together with the fixed list of parameters that a complete description of each one must contain. Core questions do not involve combinations of transformations and do not use negative scale factors, and every Core explanation, worked example, retrieval item and diagram in this chapter stays inside those limits. Extended depth is added beside each concept and never in a detached appendix: reflection in any straight line including y equals x and y equals minus x, rotation about any centre with the perpendicular bisector method for locating an unknown one, negative scale factors with the opposite ray and the area factor k squared, and combinations of transformations with the single equivalent transformation. Extended E7.2 Vectors in two dimensions covers column, directed segment and lower case bold notation, addition, subtraction and multiplication by a scalar, the reversal rule that BA is the negative of AB, and the route rule that AB plus BC equals AC, from which AB equals b minus a follows. Extended E7.3 Magnitude of a vector covers the length of a column vector as the square root of x squared plus y squared, its origin in Pythagoras, modulus notation, and exact surd or three significant figure answers. Extended E7.4 Vector geometry covers directed line segments, position vectors, expressing an unknown vector in terms of two coplanar vectors by choosing a route, midpoint vectors, division of a line in a given ratio derived rather than quoted, and proofs that vectors are parallel, that three points are collinear, that a quadrilateral is a parallelogram, and problems involving ratio and similarity. A route key, paired Core and Extended coverage maps, correct paper orientation for both routes, a visible M-O-V-E protocol, a transformation description checklist, twelve mathematically checked coordinate diagrams, route labelled worked examples, a negative enlargement clinic, a vector proof clinic, a route tagged mistake clinic, separate Core and Extended high yield summaries and exit checks, a route labelled mixed challenge and a spaced review plan support both first pass learning and last week revision.Show more

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Chapter overview

A summary of this Mathematics chapter — open a section to read it. The full notes, worked examples and practice questions are in the study modules above.

What is Transformations and Vectors about?

1. \((-3,2)\) is in the second quadrant (left and up); \((4,-1)\) is in the fourth (right and down). 2. The \(x\)-axis is \(y=0\); the \(y\)-axis is \(x=0\); the vertical line through \((2,0)\) is \(x=2\); the horizontal line through \((0,-3)\) is \(y=-3\). 3. \(\sqrt{5^{2}+12^{2}}=\sqrt{169}=13\). 4. Length scale factor \(\tfrac{10}{4}=2.5\); area scale factor \(2.5^{2}=6.25\). 5. \(3+2=5\) parts, so the first part is \(\tfrac35\) of the whole, which is 12 cm; the second is 8 cm. 6. \(y=x\).

“Rotation of \(90^\circ\) about the origin.” No direction. There are two different \(90^\circ\) rotations about the origin and this does not say which. “Reflection.” No mirror line, so the transformation is not determined at all. “Enlargement, scale factor 3.” No centre. The scale factor fixes the size of the image but not where it is. “Translation 4 right and 3 down.” Acceptable in meaning, but the syllabus asks for the vector; write \(\begin{pmatrix}4\\-3\end{pmatrix}\).

“Enlargement, centre \((1,1)\), scale factor 2” for an image on the far side of the centre. The sign is missing. If the image is on the opposite ray, the scale factor is \(-2\), and \(2\) describes a different image altogether. “Reflection in the line \(y=x\), then translation by \(\begin{pmatrix}0\\2\end{pmatrix}\).” Two transformations offered where one was asked for. Find the single equivalent transformation instead.

Core retrieval check 7.1 Core C7.1 Triangle \(A\) has vertices \((1,1)\), \((4,1)\) and \((1,3)\). Write down the vertices of its image under each of the following. (a) Reflection in the line \(x=5\). A vertical mirror leaves every \(y\)-coordinate alone and sends \(x\) to \(10-x\), because the image must be as far to the right of \(x=5\) as the object is to the left. So the image is \((9,1)\), \((6,1)\), \((9,3)\). (b) Rotation through \(90^\circ\) clockwise about the origin. The rule is \((x,y)\mapsto(y,-x)\) — check it on \((1,0)\), which must go to \((0,-1)\), and it does. So the image is \((1,-1)\), \((1,-4)\), \((3,-1)\). (c) Rotation through \(180^\circ\) about the vertex \((1,1)\). The centre is a vertex, so that vertex does not move. For the others, go from the centre to the point and then the same distance beyond: \((4,1)\) is 3 right of the centre, so its image is 3 left, at \((-2,1)\); \((1,3)\) is 2 up, so its image is 2 down, at \((1,-1)\). The image is \((1,1)\), \((-2,1)\), \((1,-1)\). No direction is needed for a half-turn. (d) Enlargement, centre \((1,1)\), scale factor \(2\). Use \(P'=C+k(P-C)\). The vertex \((1,1)\) is the centre, so it does not move. \((4,1)\): \(C+2\begin{pmatrix}3\\0\end{pmatrix}=(1,1)+(6,0)=(7,1)\). \((1,3)\): \(C+2\begin{pmatrix}0\\2\end{pmatrix}=(1,1)+(0,4)=(1,5)\). The image is \((1,1)\), \((7,1)\), \((1,5)\) — twice the size, on the same side of the centre. (e) Translation by \(\begin{pmatrix}-3\\2\end{pmatrix}\). Three left and two up: \((-2,3)\), \((1,3)\), \((-2,5)\).

Extended retrieval check 7.1 Extended E7.1 The same triangle \(A\), with vertices \((1,1)\), \((4,1)\) and \((1,3)\). (a) Reflection in \(y=-x\). The rule is \((x,y)\mapsto(-y,-x)\), so the image is \((-1,-1)\), \((-1,-4)\), \((-3,-1)\). (b) Enlargement, centre \((1,1)\), scale factor \(-2\). The vertex \((1,1)\) is the centre, so it does not move. \((4,1)\): \(C+(-2)\begin{pmatrix}3\\0\end{pmatrix}=(1,1)+(-6,0)=(-5,1)\). \((1,3)\): \(C+(-2)\begin{pmatrix}0\\2\end{pmatrix}=(1,1)+(0,-4)=(1,-3)\). The image is \((1,1)\), \((-5,1)\), \((1,-3)\) — twice the size and on the far side of the centre. Compare it with part (d) of the Core check: same centre, same \(|k|\), opposite side. (c) Triangle \(A\) is reflected in the \(x\)-axis and the image is then reflected in the \(y\)-axis. Describe fully the single transformation that has the same effect. Reflecting in \(y=0\) gives \((1,-1)\), \((4,-1)\), \((1,-3)\); reflecting that in \(x=0\) gives \((-1,-1)\), \((-4,-1)\), \((-1,-3)\). Comparing the original with the final image, both coordinates have changed sign, so the single transformation is a rotation through \(180^\circ\) about \((0,0)\). (An enlargement, centre \((0,0)\), scale factor \(-1\) sends every point to the same place, so it describes the same transformation.)

Clinic check. A shape of area \(6\,\mathrm{cm^2}\) is enlarged from a centre \(C\) by scale factor \(-\tfrac32\). (a) Is the image bigger or smaller than the object? (b) On which side of \(C\) does it lie? (c) What is its area? (a) Bigger. The size depends on \(\left|-\tfrac32\right|=\tfrac32\), so every length is one and a half times as long. (b) On the opposite side of \(C\) from the object, because \(k\) is negative. (c) Area scales by \(k^{2}=\left(-\tfrac32\right)^{2}=\tfrac94\), so the image has area \(6\times\tfrac94=13.5\,\mathrm{cm^2}\). Note that \(k^2\) is positive even though \(k\) is not — an area can never come out negative.

Key ideas to remember

The one-line version. Name the rule, give every parameter it needs, and — on the Extended route — for vectors, pick a route and say what it proves.
Derive it, do not memorise it. Two lines of route rule get you \(\overrightarrow{AB}=\mathbf b-\mathbf a\) in about four seconds, and they get it right every time. Memorising the result alone is exactly how it ends up reversed under pressure. End point minus start point is the same rule you already use for the gradient and the distance between two points.

What you need to be able to do

I can reflect a shape in a given vertical or horizontal line, and state that the mirror line is the perpendicular bisector of every point-to-image segment.
I can find an unknown vertical or horizontal mirror line from a shape and its image, and write its equation as \(x=a\) or \(y=b\).
I can rotate a shape through \(90^\circ\), \(180^\circ\) or \(270^\circ\) in a stated direction about the origin, about a vertex of the shape, or about the midpoint of one of its edges.
I can identify which of those permitted centres a given rotation used, and check my answer on a second point.
I can enlarge a shape from a centre by a scale factor that is greater than 1 or a positive fraction, and I know the image stays on the same ray from the centre.
I can say what an enlargement does to lengths, to angles and to area.
I can translate a shape by a column vector and read a translation off a diagram.
I can write a complete description of any of the four transformations, with every parameter it requires.
I know that Core questions use one transformation at a time and never a negative scale factor, so an answer that needs either is a signal that I have misread something.
I use a ruler for every straight edge I draw.
I can reflect a shape in any straight line, including \(y=x\) and \(y=-x\).
I can rotate a shape about any centre, and locate an unknown centre from two point-to-image segments using perpendicular bisectors.
I can enlarge a shape by a negative scale factor, and I know which side of the centre the image lands on and how big it is.
I can apply two transformations in a stated order, record the intermediate image, and describe the single transformation equivalent to the pair.
I can read and write a vector as a column, as \(\overrightarrow{AB}\), and as \(\mathbf a\), and use a column vector to describe a translation.
I can add and subtract column vectors and multiply one by a scalar.
I can use \(\overrightarrow{BA}=-\overrightarrow{AB}\) to reverse a directed segment.
I can build a route with \(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\) and find an equivalent route between the same two points.
I can derive \(\overrightarrow{AB}=\mathbf b-\mathbf a\) rather than memorising it, and I get the subtraction the right way round.
I can calculate \(\left|\begin{pmatrix}x\\y\end{pmatrix}\right|=\sqrt{x^{2}+y^{2}}\) and explain why it is Pythagoras.
I can leave a magnitude as an exact surd, or give it to 3 significant figures when asked for a decimal.
I can use modulus notation correctly and I never write a magnitude as a column.
I know a magnitude is a non-negative scalar, so a negative answer is a signal to check.
I can represent a vector by a directed line segment and use position vectors from an origin \(O\).
I can express an unknown vector in terms of two coplanar vectors by choosing a route.
I can find the position vector of a midpoint, and of a point dividing a line in a given ratio, by deriving it rather than quoting a formula.
I can prove two vectors are parallel by showing one is a scalar multiple of the other.
I can prove three points are collinear, and I finish the proof with an explicit sentence.
I can use vectors to settle a ratio or a similarity claim in a geometric figure.

Key terms in Transformations and Vectors

Transformations: A transformation is a rule that assigns to every point of the plane a single image point, so that a whole shape is carried to an image shape. Cambridge IGCSE Mathematics 0580 examines four: reflection, rotation, enlargement and translation. On the Core route, statement C7.1 restricts them: a reflection is in a vertical or horizontal line, a rotation is about the origin, a vertex of the shape or the midpoint of one of its edges through a multiple of 90 degrees, an enlargement uses a positive or fractional scale factor, and a translation is by a column vector; Core questions do not involve combinations of transformations. On the Extended route, statement E7.1 removes those restrictions: the mirror line may be any straight line, the centre of rotation may be any point, the scale factor may also be negative, and questions may involve combinations of transformations. A ruler must be used for all straight edges on both routes. Reflection, rotation and translation preserve every length and angle, so the image is congruent to the object; reflection alone reverses orientation. Enlargement preserves angles and shape but multiplies every length by the modulus of the scale factor and every area by its square, so the image is similar to the object. Describing a transformation completely means stating its name together with every parameter it requires: the mirror line as an equation for a reflection, the angle, direction and centre for a rotation, the centre and scale factor for an enlargement, and the column vector for a translation.
Vector Geometry: Vector geometry is the use of directed line segments and position vectors to settle geometric questions by calculation instead of by measurement. Fixing an origin O gives every point A a position vector OA, usually written a, and any vector between two points is then the difference of their position vectors, so AB equals b minus a. An unknown vector is found by choosing a route from its start point to its end point built only from vectors that are already known, since every valid route gives the same answer. The midpoint M of AB has position vector one half of a plus b, and a point P dividing AB internally in the ratio m to n has position vector n a plus m b all over m plus n, a result derived by travelling to A and then m over m plus n of the way along AB rather than quoted. Two non-zero vectors are parallel exactly when one is a scalar multiple of the other, and three points A, B and C are collinear exactly when AB is a scalar multiple of AC, because the two vectors are then parallel and share the point A. A parallelogram is established by showing one pair of opposite sides is equal as vectors, which fixes both length and direction, since equal magnitudes alone would not. The size and sign of the scalar also give relative positions and length scale factors, which is how vectors settle questions about ratio and similarity.
Vectors in Two Dimensions: A vector in two dimensions is a quantity with both magnitude and direction, recorded as a column vector whose top entry is the horizontal displacement and whose bottom entry is the vertical displacement. Cambridge prints vectors either as a directed line segment such as AB with an arrow above it, or as a lower case bold letter such as a. A vector fixes a movement but not a starting point, so two arrows of the same length and direction anywhere in the plane represent the same vector. Vectors are added and subtracted entry by entry and multiplied by a scalar entry by entry, so k times the column x, y is the column kx, ky. Reversing a directed segment reverses its vector, giving BA equal to minus AB, and following one directed segment by another gives the route rule AB plus BC equals AC. When the position vectors of A and B from an origin O are a and b, the route from A through O to B gives AB equal to b minus a, with the subtraction always in that order. A column vector is also exactly what describes a translation, which is why the same notation serves both halves of this topic.
Magnitude of a Vector: The magnitude of a vector is its length, written between modulus signs, so the magnitude of a is written mod a and the magnitude of the directed segment AB is written mod AB. For a column vector with entries x and y the magnitude is the square root of x squared plus y squared, because the two components are the perpendicular legs of a right-angled triangle whose hypotenuse is the vector itself, so the result is Pythagoras theorem applied to a displacement. A magnitude is a scalar, not a vector: it is a single non-negative number that measures how long the vector is and says nothing about which way it points. Squaring removes any minus signs, so a vector and its reverse always have the same magnitude, and the answer can never come out negative. Magnitudes may be left exact as an integer or a surd, or given as a decimal to three significant figures when a decimal is asked for. Multiplying a vector by a scalar k multiplies its magnitude by the modulus of k. The magnitude formula is not printed in the list of formulas on the examination paper, so it has to be recalled.

Common mistakes to avoid

The non-calculator paper is the lower-numbered one on each route. Paper 1 for Core, Paper 2 for Extended. It is easy to assume the calculator paper comes first because it is the harder-sounding one; it does not. Papers 3 and 4 are the calculator papers, and a scientific calculator with trigonometric functions is strongly recommended for them. Algebraic and graphical calculators are not permitted on any paper.
A seventh, quieter one Extended E7.1. Order matters when transformations are combined. Two translations are the safe exception: they always give the same result in either order, because adding column vectors is commutative. A handful of other pairs commute too — two rotations about the same centre, two enlargements from the same centre — but every one of those shares a fixed point. Where no fixed point is shared, assume reversing the order changes the image. Core questions do not combine transformations, so this danger belongs to the Extended route alone.
What a Core description can and cannot contain. On the Core route the mirror line is always \(x=a\) or \(y=b\); the centre of rotation is the origin, a vertex of the shape or the midpoint of one of its edges; and the scale factor is positive. If your description needs anything outside those, re-read the question — you have almost certainly matched the wrong pair of points. The \(180^\circ\) exception still applies: a half-turn needs no direction, because clockwise and anticlockwise land in the same place.
The \(k=-1\) overlap. A rotation of \(180^\circ\) about \(C\) puts every point exactly where an enlargement from \(C\) with \(k=-1\) puts it. They are one transformation of the plane described two ways, and each description is complete on its own. That is the one genuine overlap between the four; everywhere else exactly one description fits.
“The mirror line” is not an answer. Nor is drawing it on the diagram without naming it. The mark is for the equation. If the mirror is the vertical line through \((3,0)\), write \(x=3\) — not \(y=3\), which is a different line entirely.
A fractional scale factor is still an enlargement. “Enlargement, centre \((2,2)\), scale factor \(\tfrac13\)” is the correct description of a shape that has become smaller; there is no separate word for it in this syllabus, and writing “reduction” does not answer the instruction. Note also that the centre itself is the one point that never moves, whatever \(k\) is.
Size and sign are different questions. The sign of \(k\) decides which side of the centre the image is on; \(|k|\) decides how big it is. So \(k=-3\) makes the image three times as large and puts it on the opposite side, while \(k=\tfrac12\) makes it half as large on the same side. Reading “negative” as “smaller” conflates the two and answers the wrong question twice over. The negative enlargement clinic works through all four cases from one centre.
Top row across, bottom row up. Writing \(\begin{pmatrix}3\\-2\end{pmatrix}\) when you meant 3 down and 2 left is not a small slip; it is a different transformation. Three down and two left is \(\begin{pmatrix}-2\\-3\end{pmatrix}\). Read back what you have actually written — \(\begin{pmatrix}3\\-2\end{pmatrix}\) says “three across, two down” — and check each component against the diagram before moving on.
Why the lines through \(P,P'\) all meet at the centre. Because \(\overrightarrow{CP'}=k\overrightarrow{CP}\), the points \(C\), \(P\) and \(P'\) are always collinear — whatever the value of \(k\). So every line joining an object vertex to its image passes through the centre, and two of them are enough to find it. This is the same fact as the dashed rays in Figures 7.4 and 7.5, used in reverse.
A point is not a vector. \(A\) is a place. \(\overrightarrow{OA}\) is the journey from the origin to that place, and \(\mathbf a\) is a name for that journey. They happen to carry the same pair of numbers when the journey starts at \(O\), which is why the two are easily confused — but \(\overrightarrow{AB}\) does not start at \(O\), and its components are differences, not coordinates. In handwriting, underline a vector letter (\(\underline{a}\)); print uses bold.
\(\mathbf a-\mathbf b\) and \(\overrightarrow{AB}\) are not the same thing. In the example above \(\mathbf a-\mathbf b=\begin{pmatrix}6\\-2\end{pmatrix}\) but \(\overrightarrow{AB}=\begin{pmatrix}-6\\2\end{pmatrix}\). They differ by a sign, because \(\overrightarrow{AB}=\mathbf b-\mathbf a\) subtracts the start from the end. Whenever the letters in a directed segment run \(A\) to \(B\), the vector runs \(\mathbf b\) minus \(\mathbf a\) — the reverse of the reading order, which is exactly why it catches people out.
Three things a magnitude can never be. It can never be a column — \(\left|\begin{pmatrix}-5\\12\end{pmatrix}\right|\) is \(13\), not \(\begin{pmatrix}5\\12\end{pmatrix}\). It can never be negative, so a minus sign in an answer means a slip in the squaring. And it can never be the sum of the components: \(-5+12=7\) is not the length of anything here. If your answer to a magnitude question is not a single non-negative number, stop and re-read the working.
Both vectors must start at the same point. Showing that \(\overrightarrow{MN}\) is a multiple of \(\overrightarrow{NP}\) also works, because those two share \(N\). But showing \(\overrightarrow{MN}\) is a multiple of some unrelated vector proves only that two lines are parallel — parallel lines need never meet, so no collinearity follows. The shared point is doing half the work in every one of these proofs, and it must be named in the conclusion.
Equal magnitudes are not enough. \(\left|\begin{pmatrix}4\\1\end{pmatrix}\right| =\left|\begin{pmatrix}1\\4\end{pmatrix}\right|=\sqrt{17}\), but those two vectors point in completely different directions. A quadrilateral with one pair of opposite sides merely equal in length could be an isosceles trapezium, not a parallelogram. The vector statement \(\overrightarrow{AB}=\overrightarrow{DC}\) is stronger than the length statement \(|AB|=|DC|\), and it is the stronger one you need.
Mind the letter order. For the quadrilateral \(ABCD\), the side opposite \(AB\) is \(DC\), not \(CD\) — because going round the quadrilateral is \(A\to B\to C\to D\), so \(AB\) and \(CD\) run in opposite senses. Writing \(\overrightarrow{AB}=\overrightarrow{CD}\) claims something false; the correct statement is \(\overrightarrow{AB}=\overrightarrow{DC}\), or equivalently \(\overrightarrow{AB}=-\overrightarrow{CD}\).
The rotation rules are for the origin only. For a rotation about a vertex or an edge midpoint, work with the journey from that centre instead: find \(\overrightarrow{CP}\), turn it through the angle, then add it back to \(C\). Applying an origin rule to a non-origin centre gives the wrong image every time — and two of the three Core centres are not the origin.

Examiner tips

Route meaning never depends on colour. Every badge carries the word Core or Extended and the exact syllabus reference, and every Extended block carries a written label, so the boundary survives greyscale printing and a black-and-white photocopy.
The Core objective to check first is the eighth one: writing a complete description. Whenever a question says “describe fully”, it is asking for a fixed list of parameters, and supplying that list costs nothing beyond the work you have already done. It is the one objective on this page that can be secure in the mathematics and still missing on the page.
The Extended objective to check first is the last one in E7.4. The syllabus asks you to show that vectors are parallel and to show that three points are collinear; an equation on its own has not shown anything yet. The concluding sentence is part of the answer, not a courtesy.
Read the notes column, not just the skill. Every difference between the two routes in Topic 7 lives in the syllabus notes rather than in the numbered statements: which mirror lines, which centres of rotation, which scale factors, and whether combinations appear. The skill sentences are nearly identical; the limits are not. That is exactly why a Core candidate who practises on Extended material wastes time, and why an Extended candidate who practises only on Core material is under-prepared.
On rounding. The general 0580 convention applies here as everywhere: where answers are not exact, give them to 3 significant figures unless the question defines a different accuracy; answers that are exact to four or five significant figures are not rounded; and carry a higher accuracy through the working rather than rounding an intermediate value. Most of Topic 7 produces exact answers anyway, so the convention bites mainly on Extended magnitudes.
M-O-V-E is a specialisation of R-I-S-E, the whole-course protocol — read and represent, identify connections, solve visibly, evaluate. If you already use R-I-S-E, keep it; M-O-V-E just names what “represent” and “evaluate” mean when the objects are shapes and directed segments.
Check a coordinate rule on one easy point rather than trusting your memory. Take \((1,0)\). Reflected in \(y=x\) it must land on \((0,1)\); reflected in \(y=-x\) it must land on \((0,-1)\). If a rule you have written does not do that, it is the wrong rule, and the check costs two seconds.
Tracing paper is available on all four papers, and you may ask for it during the examination. Trace the object, put your pencil point on a candidate centre, and turn the paper through the angle: if the tracing lands on the image, the centre and angle are right. Use it to confirm a centre you found by the method above, not to replace the working — the description still has to be written out in full.
The pair you can rely on. Two translations always give the same result in either order, because \(\mathbf u+\mathbf v=\mathbf v+\mathbf u\). A few other pairs happen to commute as well — two rotations about the same centre, or two enlargements from the same centre — but every one of those shares a fixed point, which is a condition to check rather than assume. Every pairing that does not, such as reflection then rotation or rotation then enlargement, should be assumed to depend on the order unless you have checked otherwise.
Simplify to the standard form. Leave an answer as a tidy combination such as \(\tfrac12\mathbf p+\tfrac12\mathbf r\) or \(\mathbf r-\mathbf p\), with like terms collected. An unsimplified expression such as \(\mathbf p+\tfrac12\mathbf r-\tfrac12\mathbf p\) is not yet an answer in the form the question asked for, and it hides the scalar multiple you may need in the next part.
Not on the formula sheet. The Extended list of formulas printed on page 2 of Papers 2 and 4 covers areas, circumference, curved surface areas and volumes, the quadratic formula, and the sine rule, cosine rule and area of a triangle. Nothing from Topic 7 appears on it, so vector magnitude is a recall item. Since it is just Pythagoras, the safest way to “remember” it is to sketch the right triangle.
The last column is not optional prose. The syllabus asks you to show that vectors are parallel and to show that three points are collinear, and an equation on its own has not yet shown anything — it is the working that leads to the statement. Reaching the correct scalar multiple and stopping leaves the question unanswered. One sentence, every time.
Read the question before choosing. If it says “describe fully the single rotation”, give the rotation. If it says “describe fully the single enlargement”, give \(k=-1\). If it just says “describe fully”, pick one and give every parameter it needs, rather than hedging with both.
Six of these eight cost no mathematics at all. Errors 1, 2, 4, 6, 7 and 8 are failures of recording, not of understanding — the work was done and the answer was written down badly. All three of the Core errors on this page are in that group, so fixing them needs no new mathematics at all — it needs the habit line under each one. Rereading the list before you sit down to a past paper takes about ten minutes.
What the Core formula list actually contains. The list printed on page 2 of Papers 1 and 3 covers the area of a triangle and of a circle, the circumference of a circle, the curved surface areas of a cylinder and a cone, the surface area of a sphere, and the volumes of a prism, pyramid, cylinder, cone and sphere. Nothing from Topic 7 is on it, so the enlargement relation and the area factor are both recall items.
What the Extended formula list actually contains. The list printed on page 2 of Papers 2 and 4 is the Core list plus the quadratic formula and the sine rule, cosine rule and \(\tfrac12ab\sin C\). Nothing from Topic 7 is on it, so all seven relationships above are recall items. The good news is that six of the seven can be rebuilt in a line or two from the route rule and Pythagoras, so what you really need to hold is the route rule, the enlargement relation, and the habit of deriving the rest.
Mark yourself against the descriptions, not just the numbers. In questions 1, 4, 5 and 9 the coordinates are worth about half the marks and the complete description is worth the rest. If you got every image right and still scored under half, the problem is not your mathematics — go back to the description checklist.

How Transformations and Vectors is examined

Every candidate takes exactly two components, and which two depends on the route. Both of your papers may assess Topic 7, and on both of them a ruler is needed — the syllabus states explicitly that a ruler must be used for all straight edges in transformation work.
The non-calculator paper is the lower-numbered one on each route. Paper 1 for Core, Paper 2 for Extended. It is easy to assume the calculator paper comes first because it is the harder-sounding one; it does not. Papers 3 and 4 are the calculator papers, and a scientific calculator with trigonometric functions is strongly recommended for them. Algebraic and graphical calculators are not permitted on any paper.
Ruler, compasses, protractor. Candidates on both routes are expected to have all three for every paper. In Topic 7 the ruler is the one that matters: the syllabus requires it for all straight edges, so rule them even when the vertices are already plotted correctly.
Tracing paper is available on all four papers. You cannot bring your own, but you may request it during the examination. It is genuinely useful for checking a rotation, and it is the fastest way to confirm a centre you have already found.
Answers are written on the question paper, and all necessary working must be shown in the spaces provided.
Label the image. A drawn transformation that is not labelled leaves the examiner guessing which shape is your answer.

Syllabus reference and sources

Written against: Cambridge IGCSE Mathematics (0580) syllabus for examination in 2025, 2026 and 2027, version 3 (Subject Content, Topic 7: Transformations and vectors — Core row C7.1 and Extended rows E7.1 to E7.4).

Written by: Academiq Edu Instructor Panel

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