Transformations
- Describe and perform translations using column vectors
- Describe and perform reflections in any mirror line including $y=x$ and $y=-x$
- Describe and perform rotations, stating centre, angle, and direction
- Describe and perform enlargements including fractional and negative scale factors
- Find the single transformation equivalent to a combination of two transformations
🔑 Core Concepts
1. Translation
A translation moves every point of a shape by the same distance in the same direction. It is the only transformation that does not involve a fixed point. The shape is neither rotated nor reflected — only shifted. Translations are described using a column vector.
2. Reflection
A reflection maps each point to its mirror image across a line of symmetry. The perpendicular distance from each point to the mirror line equals the perpendicular distance from the image to the mirror line. Reflections reverse orientation (a clockwise shape becomes anti-clockwise in its image).
3. Rotation
A rotation turns a shape about a fixed point called the centre of rotation. Every point moves through the same angle. A rotation is an isometry — it preserves size and shape but changes orientation (unless it is 180°). To fully describe a rotation you need three things: the centre, the angle, and the direction.
Finding the Centre of Rotation: Draw the perpendicular bisector of the line segment from each vertex to its image. The centre of rotation is the intersection of these perpendicular bisectors. Algebraically, the centre $(a, b)$ is equidistant from each point and its image when measured along an arc.
4. Enlargement
An enlargement changes the size of a shape (and can reverse orientation when the scale factor is negative). Lengths are multiplied by the scale factor but angles remain unchanged. An enlargement is not an isometry unless $|k|=1$.
5. Combined Transformations & Invariant Points
Two or more transformations applied in sequence are called combined transformations. A key Grade 9 skill is identifying the single equivalent transformation. The order matters: applying transformation $A$ then $B$ is generally not the same as $B$ then $A$.
🗺️ Visual Notes
- Column vector $\binom{a}{b}$
- $(x,y)\to(x+a,y+b)$
- Preserves size & orientation
- No invariant points
- State the mirror line equation
- $y=x$: swap coords
- $y=-x$: swap & negate
- Reverses orientation
- State centre, angle, direction
- 90° CW: $(x,y)\to(y,-x)$
- Preserves size & shape
- Centre is only invariant point
- State SF and centre
- Negative SF: image opposite side
- Fractional SF: image smaller
- Area SF $= k^2$
- Order matters
- Find single equivalent
- Check orientation preserved?
- Identify invariant points
- Negative SF enlargements
- Centre of rotation algebraically
- Single equivalent transformation
- Invariant points
Transformation Comparison Table
| Transformation | Preserves Size? | Preserves Orientation? | Invariant Points | Full Description Requires |
|---|---|---|---|---|
| Translation | Yes | Yes | None | Column vector only |
| Reflection | Yes | No (reversed) | All points on mirror line | Equation of mirror line |
| Rotation | Yes | Yes (unless 180°) | Centre only | Centre, angle, direction |
| Enlargement (SF $>0$) | No | Yes | Centre only (if SF≠1) | Scale factor, centre |
| Enlargement (SF $<0$) | No | No (reversed) | Centre only | Scale factor (neg.), centre |
Reflection Lines Quick Reference
| Mirror Line | Equation | Rule | Example: $(3, 5)$ maps to |
|---|---|---|---|
| $x$-axis | $y=0$ | $(x,y)\to(x,-y)$ | $(3,-5)$ |
| $y$-axis | $x=0$ | $(x,y)\to(-x,y)$ | $(-3,5)$ |
| Diagonal (up-right) | $y=x$ | $(x,y)\to(y,x)$ | $(5,3)$ |
| Diagonal (up-left) | $y=-x$ | $(x,y)\to(-y,-x)$ | $(-5,-3)$ |
| Vertical line | $x=a$ | $(x,y)\to(2a-x,y)$ | e.g. $x=4$: $(5,5)$ |
| Horizontal line | $y=b$ | $(x,y)\to(x,2b-y)$ | e.g. $y=2$: $(3,-1)$ |
Decision Tree: Identifying a Transformation
Find SF and centre
Find mirror line
Find centre, angle, direction
Find column vector
✏️ Worked Examples
$B(3,2) \to (3+(-4),\; 2+3) = (-1, 5)$
$C(2,4) \to (2+(-4),\; 4+3) = (-2, 7)$
$B(3,2) \to (-3, 2)$
$C(2,4) \to (-2, 4)$
(a) Image vertices: $A'(-3,5)$, $B'(-1,5)$, $C'(-2,7)$
(b) Image vertices: $A''(-1,2)$, $B''(-3,2)$, $C''(-2,4)$
$Q(4,1) \to (-8,-2)$
$R(4,3) \to (-8,-6)$
$S(1,3) \to (-2,-6)$
Description: Enlargement, scale factor $-2$, centre $(0,0)$.
$B(5,1)\to B'(1,5)$
$C(5,4)\to C'(4,5)$
$B'(1,5)\to B''(-5,1)$
$C'(4,5)\to C''(-5,4)$
Notice: $x$-coordinate negated, $y$-coordinate unchanged. The rule is $(x,y)\to(-x,y)$, which is reflection in the $y$-axis.
(b) Reflection in the $y$-axis (equation $x=0$)
(c) All points on the $y$-axis, i.e. $(0, k)$ for any $k \in \mathbb{R}$
❓ Exam Questions
Write the column vector that translates the point $(5, -3)$ to the point $(2, 4)$.
Mark scheme: B1 for $\binom{-3}{7}$ (or equivalent correct vector notation). Calculation: $2-5=-3$ and $4-(-3)=7$.
Describe fully the single transformation that maps triangle $P$ with vertices $(1,2)$, $(3,2)$, $(3,5)$ to triangle $Q$ with vertices $(-1,2)$, $(-3,2)$, $(-3,5)$.
Mark scheme: M1 for identifying it as a reflection; A1 for stating the correct mirror line $x=0$ (or "$y$-axis"). Just writing "reflection" without the line scores M1 only.
Describe fully the single transformation that maps shape $A$ with vertices $(2,1)$, $(6,1)$, $(6,3)$ to shape $A'$ with vertices $(-1,-2)$, $(-1,-6)$, $(-3,-6)$.
Working: Check the rule $(x,y)\to(y,-x)$: $(2,1)\to(1,-2)$ ✓; $(6,1)\to(1,-6)$ ✓; $(6,3)\to(3,-6)$ ✓.
Mark scheme: B1 rotation; B1 90° clockwise (or 270° anti-clockwise); B1 centre $(0,0)$. All three required for full marks.
Triangle $T$ has vertices at $(1,1)$, $(3,1)$, $(3,4)$. It is enlarged by scale factor $-\frac{1}{2}$ about the point $(1,1)$. Find the coordinates of the image vertices and describe what has happened to the triangle.
$(1,1)\to(1+(-\tfrac{1}{2})(0),\;1+(-\tfrac{1}{2})(0))=(1,1)$ — the centre maps to itself.
$(3,1)\to(1+(-\tfrac{1}{2})(2),\;1+(-\tfrac{1}{2})(0))=(1-1,1)=(0,1)$
$(3,4)\to(1+(-\tfrac{1}{2})(2),\;1+(-\tfrac{1}{2})(3))=(0,\,-\tfrac{1}{2})$
Image vertices: $(1,1)$, $(0,1)$, $(0,-\frac{1}{2})$.
The triangle is half the size, on the opposite side of the centre $(1,1)$, and rotated 180°.
Mark scheme: M1 method for applying enlargement about a non-origin centre; A1 for correct image of at least two vertices; A1 all three correct; B1 description (smaller, opposite side / equivalent).
Shape $S$ with vertices $A(1,0)$, $B(3,0)$, $C(3,2)$, $D(1,2)$ is reflected in the line $y=x$ to give $S'$. $S'$ is then reflected in the line $y=-x$ to give $S''$. (a) Find the coordinates of $S''$. (b) Describe the single transformation mapping $S$ to $S''$. (c) State the equations of any invariant lines of this single transformation.
Reflection in $y=x$: $(x,y)\to(y,x)$:
$A(1,0)\to A'(0,1)$; $B(3,0)\to B'(0,3)$; $C(3,2)\to C'(2,3)$; $D(1,2)\to D'(2,1)$
Reflection in $y=-x$: $(x,y)\to(-y,-x)$:
$A'(0,1)\to A''(-1,0)$; $B'(0,3)\to B''(-3,0)$; $C'(2,3)\to C''(-3,-2)$; $D'(2,1)\to D''(-1,-2)$
(b) Compare original to $S''$: $A(1,0)\to(-1,0)$; $B(3,0)\to(-3,0)$; $C(3,2)\to(-3,-2)$; $D(1,2)\to(-1,-2)$.
Rule: $(x,y)\to(-x,-y)$. This is a rotation of 180° about the origin.
(c) Any line through the origin is invariant under 180° rotation (each point maps to a different point on the same line through origin — but the line itself is mapped to itself). So all lines through $(0,0)$ are invariant lines, e.g. $y=mx$ for any $m$.
Mark scheme: A1 each for $A''$, $B''$; A1 all four correct (a); M1 comparing original and final, A1 rotation 180° about origin (b); B1 lines through the origin/any line $y=mx$ (c).
⭐ Grade 9 Model Answers
Full annotated answer to Q5 (the 6-mark combined transformations question):
Part (a): Finding $S''$
[Mark 1: Correct application of $y=x$ reflection — "I know the rule $(x,y)\to(y,x)$ for $y=x$ because the $y=x$ line swaps coordinates."]
$A(1,0)\to A'(0,1)$, $B(3,0)\to B'(0,3)$, $C(3,2)\to C'(2,3)$, $D(1,2)\to D'(2,1)$ ✓
[Mark 2–3: Correct application of $y=-x$ reflection — "The rule for $y=-x$ is $(x,y)\to(-y,-x)$: swap AND negate both."]
$A'(0,1)\to(-1,0)$, $B'(0,3)\to(-3,0)$, $C'(2,3)\to(-3,-2)$, $D'(2,1)\to(-1,-2)$ ✓ (all four correct earns A1)
Part (b): Single equivalent transformation
[Mark 4–5: A model answer always checks multiple points, not just one.]
Comparing $S$ to $S''$: the rule is $(x,y)\to(-x,-y)$. This is recognised as 180° rotation about the origin. I verify: rotation 180° about $O$ maps $(x,y)\to(-x,-y)$ ✓. I state direction? For 180°, no direction needed — clockwise and anti-clockwise give the same result.
Answer: Rotation, 180°, about the origin $(0,0)$.
Part (c): Invariant lines
[Mark 6: Grade 9 mark — requires conceptual understanding, not just calculation.]
An invariant line is a line that maps to itself (not necessarily point by point). Under 180° rotation about $O$, a point $(a,b)$ on the line $y=mx$ maps to $(-a,-b)$. Is $(-a,-b)$ also on $y=mx$? Yes: $-b = m(-a) \Rightarrow b=ma$ ✓. So every line $y=mx$ through the origin is invariant.
Answer: Any line of the form $y=mx$ (all lines through the origin) is an invariant line.
- Part (a): Shows all intermediate steps; doesn't skip the $S'$ stage. Shows working for each vertex separately.
- Part (b): Identifies the rule algebraically, then matches it to a named transformation. States all three required components (rotation, 180°, origin).
- Part (c): Uses the definition of invariant line (line maps to itself) rather than invariant point. Proves it rather than just stating it.
📋 Revision Sheet
| Term | Meaning |
|---|---|
| Translation | Slide — no rotation/reflection; uses column vector |
| Reflection | Mirror flip; reverses orientation |
| Rotation | Turn about a centre point |
| Enlargement | Scale change from a centre point |
| Isometry | Transformation preserving size (translation, reflection, rotation) |
| Invariant point | Point that maps to itself |
| Invariant line | Line that maps to itself (as a set) |
$\text{Translation }\binom{a}{b}$: $(x,y)\to(x+a,y+b)$
Reflection $x$-axis: $(x,y)\to(x,-y)$
Reflection $y$-axis: $(x,y)\to(-x,y)$
Reflection $y=x$: $(x,y)\to(y,x)$
Reflection $y=-x$: $(x,y)\to(-y,-x)$
Rot 90°CW about $O$: $(x,y)\to(y,-x)$
Rot 90°ACW about $O$: $(x,y)\to(-y,x)$
Rot 180° about $O$: $(x,y)\to(-x,-y)$
Enlargement SF $k$ about $O$: $(x,y)\to(kx,ky)$
Enlargement SF $k$ about $(a,b)$: $(x,y)\to(a+k(x-a), b+k(y-b))$
Area SF $= k^2$ for enlargement SF $k$
- $y=x$: "SWAP" — just swap the two coordinates
- $y=-x$: "SWAP & NEGATE" — swap then make both negative
- 90°CW: "x gets y's value, y gets negated x"
- 90°ACW: "y gets x's value (negated), x gets y"
- Negative SF: "Opposite side & flip" — like 180° rotation of the enlarged shape
- TRRR: The four transformations — Translation, Reflection, Rotation, enlaRgement
- Full description checklist: Translation → vector; Reflection → line; Rotation → centre, angle, direction; Enlargement → SF, centre
- Always name the transformation type first
- Rotation: 180° needs no direction stated
- Enlargement: always state whether SF is positive or negative
- Finding centre of rotation: draw perpendicular bisectors of AA′, BB′
- Combined transformations: work through each stage; do not skip
- Area scale factor is $k^2$, not $k$ — common loss of marks
- Invariant points vs invariant lines: a line can be invariant even if its individual points are not fixed
- When asked to "describe fully" — one missing component = lose a mark
🔄 Flashcards
Click a card to reveal the answer.
✗ Common Mistakes
What students do: Write "rotation of 90° about the origin" without stating the direction.
Why marks are lost: Clockwise and anti-clockwise rotations of 90° produce completely different images. A missing direction loses 1 mark on any "describe the transformation" question.
How to avoid it: Use a mental checklist: centre, angle, direction. For 90° and 270°, direction is always required. Only for 180° can you omit direction.
What students do: When asked "how many times larger is the area of the image?", they give the scale factor $k$ instead of $k^2$.
Why marks are lost: The area scale factor is always $k^2$ (and volume is $k^3$). Writing SF $=3$ means area is 9 times larger, not 3 times.
How to avoid it: Always square the scale factor for areas. "Lengths times $k$, areas times $k^2$, volumes times $k^3$."
What students do: Apply the $y=x$ rule (swap) when the question asks for $y=-x$ (swap and negate), or vice versa.
Why marks are lost: All coordinate answers are wrong; the image is in a completely different position.
How to avoid it: Sketch the lines. $y=x$ goes from bottom-left to top-right; $y=-x$ goes from top-left to bottom-right. Check a point: $(2,5)$ reflected in $y=x$ gives $(5,2)$; in $y=-x$ gives $(-5,-2)$.
What students do: When performing an enlargement with negative SF, they correctly scale the distances but forget to place the image on the opposite side of the centre of enlargement.
Why marks are lost: The image ends up on the wrong side of the centre — all coordinates wrong.
How to avoid it: Draw a ray from the centre through each vertex. For negative SF $-k$, the image vertex is at distance $k$ times the original distance, but in the opposite direction along the ray (past the centre).
What students do: Describe a translation as "3 right and 2 up" or write it as $(3, 2)$.
Why marks are lost: The required form is always the column vector $\binom{3}{2}$. "3 right, 2 up" scores 0 marks for the description; $(3,2)$ is ambiguous and typically scores 0.
How to avoid it: Always write $\binom{a}{b}$ with one number above the other. Practise drawing the column vector notation until it becomes automatic.
What students do: Apply $(kx, ky)$ when the centre of enlargement is not the origin — forgetting to translate to the origin first.
Why marks are lost: All image coordinates are wrong. This is an extremely common error in higher-tier questions.
How to avoid it: Use the formula $(a+k(x-a), b+k(y-b))$ for centre $(a,b)$, or alternatively: subtract the centre, multiply by $k$, add the centre back. Check by verifying the centre maps to itself.
✅ Final Checklist
Click each item to mark it complete. Your progress is saved automatically.
- I can translate a shape using a column vector and write the image coordinates.
- I can describe a translation fully using correct column vector notation $\binom{a}{b}$.
- I can reflect a shape in the $x$-axis, $y$-axis, $y=x$, and $y=-x$.
- I can reflect a shape in the line $x=a$ or $y=b$ for any value $a$ or $b$.
- I can find the equation of a mirror line given a shape and its reflection.
- I can rotate a shape 90° CW, 90° ACW, and 180° about the origin using coordinate rules.
- I can find the centre of rotation using perpendicular bisectors.
- I can describe a rotation fully: centre, angle, direction.
- I can perform an enlargement with a positive, fractional, or negative scale factor about any centre.
- I can describe an enlargement fully: scale factor and centre.
- I know that area scale factor $= k^2$ for enlargement with scale factor $k$.
- I can perform two transformations in sequence and find the single equivalent transformation.
- I can identify invariant points and invariant lines for a given transformation.
- I understand negative scale factor enlargements: image is on the opposite side of the centre and is inverted.
- I can systematically describe any transformation fully without missing required components.