Learn to draw, calculate, recognise and describe transformations of shapes on coordinate grids, including negative and fractional enlargements and combined transformations.
Subtopic 1
A transformation moves or changes a shape using a defined rule. The original shape is called the object; the result is called the image. If the object has vertices \(A,B,C\), we label the image vertices \(A',B',C'\), matching corresponding points.
To map a shape means to send each of its points to a new point by the transformation. A feature that remains unchanged is invariant.
On coordinate grids, describe positions using ordered pairs \((x,y)\): first move horizontally, then vertically. Drawing an accurate grid makes transformation questions easier.
Triangle \(PQR\) becomes \(P'Q'R'\) after every vertex moves four units right. Name the object and image.
\(PQR\) is the object and \(P'Q'R'\) is the image. Each corresponding vertex is four units right of the original: the transformation is a translation. All lengths and angles are unchanged.
A rectangle is enlarged by scale factor 3. Its perimeter was 14 cm and area 12 cm². Which properties stay invariant?
The angles and shape stay unchanged; length, perimeter and area do not. Perimeter becomes \(3\times14=42\) cm and area becomes \(3^2\times12=108\text{ cm}^2\).
Subtopic 2
A reflection flips a shape across a mirror line. For each point, its image lies the same perpendicular distance on the opposite side of that line. The mirror line is the perpendicular bisector of the segment joining corresponding points.
Reflection preserves distances and angles, but reverses orientation: a clockwise order of labelled vertices becomes anticlockwise. Points on the mirror line stay fixed.
For a line \(x=a\), use \((x,y)\mapsto(2a-x,y)\). For \(y=b\), use \((x,y)\mapsto(x,2b-y)\).
Reflect \(A(3,-2)\) in the \(y\)-axis.
Only the \(x\)-coordinate changes sign.
Reflect \(B(-1,5)\) in the line \(y=2\).
The point is 3 units above the mirror line, so its image must be 3 units below it:
\[y'=2(2)-5=-1\]\[\boxed{B'=(-1,-1)}\]Triangle \(ABC\) has vertices \(A(1,2),B(3,2),C(2,5)\). Reflect it in \(y=-x\), giving the image coordinates.
Reflect each vertex by swapping the coordinates and reversing both signs:
\[A'(-2,-1),\quad B'(-2,-3),\quad C'(-5,-2)\]Plot these three points and join them in the original vertex order. All side lengths are preserved.
Drawing method: mark corresponding perpendicular distances for every corner, then join the image vertices with a ruler. For a diagonal mirror line, using tracing paper helps check the reversal.
Subtopic 3
A rotation turns a figure about a fixed centre of rotation. You must know angle, direction and centre. Distances from the centre and the size and orientation of the shape remain unchanged.
Clockwise and anticlockwise rotations are measured as viewed on the page. A quarter-turn is \(90^\circ\), a half-turn \(180^\circ\), and three quarter-turns \(270^\circ\). For example, \(270^\circ\) anticlockwise is equivalent to \(90^\circ\) clockwise.
For a different centre \((h,k)\), subtract \((h,k)\) from the point, rotate that displacement, then add \((h,k)\) back.
Rotate \(A(2,1)\) through \(90^\circ\) anticlockwise about \((0,0)\).
Rotate \(B(6,1)\) through \(180^\circ\) about \(C(2,3)\).
The centre is the midpoint between each point and its image.
\[B'=(2\times2-6,\;2\times3-1)\]\[\boxed{B'=(-2,5)}\]Rotate triangle \(ABC\) with \(A(3,1),B(5,1),C(3,4)\) through \(90^\circ\) clockwise about \((1,2)\).
Subtract the centre: \(A:(2,-1),B:(4,-1),C:(2,2)\). A clockwise quarter-turn changes \((u,v)\) into \((v,-u)\).
\[\begin{aligned}A':\;(-1,-2)+(1,2)&=\boxed{(0,0)}\\B':\;(-1,-4)+(1,2)&=\boxed{(0,-2)}\\C':\;(2,-2)+(1,2)&=\boxed{(3,0)}\end{aligned}\]Drawing method: mark the centre, count the horizontal and vertical displacement of each corner, rotate that displacement, then join corresponding image points. A rotation is not a reflection: the vertex order is preserved.
Subtopic 4
A translation slides a figure. Every point moves by the same distance in the same direction. The image has the same size and orientation as the object.
A translation can be written as a column vector: the top number is the horizontal movement, the bottom is the vertical movement. Positive values mean right/up; negative values mean left/down.
To find the translation that takes one vertex to its corresponding image, calculate \((\Delta x,\Delta y)=(x'-x,y'-y)\). Check the same vector works for the other vertices.
Translate \(P(-2,5)\) by \(\begin{pmatrix}4\-3\end{pmatrix}\).
A point \(A(-3,2)\) becomes \(A'(1,-4)\). Find the translation.
Triangle \(ABC\) has \(A(0,1),B(3,1),C(1,4)\). Translate it by \(\begin{pmatrix}-2\5\end{pmatrix}\), then by \(\begin{pmatrix}6\-8\end{pmatrix}\). Find the single equivalent translation and the final vertices.
Add the movement components:
\[\begin{pmatrix}-2\\5\end{pmatrix}+\begin{pmatrix}6\\-8\end{pmatrix}=\boxed{\begin{pmatrix}4\\-3\end{pmatrix}}\]Use this once on each original vertex:
\[\boxed{A'(4,-2),\ B'(7,-2),\ C'(5,1)}\]Subtopic 5
An enlargement changes the size of a shape from a fixed centre of enlargement using a scale factor \(k\). Lengths are multiplied by \(|k|\); angles stay the same. Normally the new shape is similar rather than congruent.
The centre itself is unchanged. To draw an image, draw rays from the centre through every vertex (extend the lines for negative factors), mark the correct distance for each image vertex, then join.
In this formula the centre is \((h,k_0)\) and scale factor is \(k\). The subscripts just distinguish the centre’s second coordinate from the scale factor. An equivalent vector form is \(P'=C+k(P-C)\).
Scale factors: length ratio \(=|k|\); area ratio \(=k^2\). A factor of \(-2\) makes side lengths twice as large and area four times as large.
Enlarge \(A(2,3)\) by scale factor \(3\) about the origin.
Enlarge \(B(5,4)\) by scale factor \(2\) about \(C(1,2)\).
First find the displacement from the centre: \((5-1,4-2)=(4,2)\). Double it and add back the centre.
\[B'=(1,2)+2(4,2)=\boxed{(9,6)}\]For a factor such as \(\tfrac12\), the image vertex is half as far from the centre along the same ray. Corresponding image sides are half as long and the area is one-quarter of the original.
An enlargement has centre \((2,2)\) and scale factor \(\tfrac12\). Find the image of \(D(8,6)\).
A negative factor places the image across the centre. For example, \(-\tfrac12\) produces a smaller image across the centre; \(-3\) produces an image three times as large across the centre. Do not confuse a negative factor with a fractional factor.
Triangle \(ABC\) has \(A(1,1),B(3,1),C(2,4)\). Enlarge it about \((2,2)\) by factor \(-2\). Find the image vertices and the area scale factor.
Lengths multiply by 2; areas by \((-2)^2=\boxed{4}\). Both original and image triangles have the same corresponding angles.
A triangle’s vertex \(A(1,2)\) maps to \(A'(3,4)\) and \(B(2,2)\) maps to \(B'(5,4)\) under an enlargement. Find the scale factor and centre.
The vector from A to B is \((1,0)\), but from \(A'\) to \(B'\) it is \((2,0)\), so \(k=2\).
Use \(A'=C+2(A-C)=2A-C\), hence \(C=2A-A'\).
\[C=(2,4)-(3,4)=\boxed{(-1,0)}\]Subtopic 6
When you are given an object and its image and asked to describe fully the single transformation, identify the type and give the extra information needed. Writing only “rotation” or “enlargement” is incomplete.
| Transformation | You must state |
|---|---|
| Reflection | Mirror line, e.g. \(y=2\) or \(y=x\) |
| Rotation | Angle, direction and centre |
| Translation | Column vector |
| Enlargement | Scale factor and centre |
Recognising by sight: if a tracing must be flipped, look for a reflection. If it must turn without flipping, look for a rotation. If it slides unchanged, look for a translation. If sizes differ, look for an enlargement.
For reflection, a mirror line passes through the midpoints of corresponding vertices at right angles. For rotation, the centre is equidistant from each point and its image. For enlargement, lines through pairs of corresponding vertices meet at the centre.
A point \(P(2,3)\) moves to \(P'(6,1)\), and all other vertices move in exactly the same way. Describe the transformation fully.
Translation by \(\boxed{\begin{pmatrix}4\\-2\end{pmatrix}}\).
\(A(-1,4)\) maps to \(A'(-1,-2)\) and \(B(3,5)\) maps to \(B'(3,-3)\). Describe the reflection.
Vertical midpoints: \(\frac{4+(-2)}2=1\) and \(\frac{5+(-3)}2=1\). All \(x\)-coordinates stay fixed.
It is a \(\boxed{\text{reflection in }y=1}\).
Corresponding vertices \(A(4,2)\mapsto A'(-2,5)\) and \(B(2,2)\mapsto B'(1,5)\) belong to a pair of similar triangles. Describe the enlargement.
The original horizontal edge goes 2 units left; its image goes 3 units right, so the scale factor is \(k=-\tfrac32\).
For centre \(C\), \(A'=C+k(A-C)\), so \(C=(A'-kA)/(1-k)\).
\[C=\frac{(-2,5)+\tfrac32(4,2)}{\tfrac52}=\frac{(4,8)}{2.5}=\boxed{(1.6,3.2)}\]Enlargement, scale factor \(\boxed{-\tfrac32}\), centre \(\boxed{(1.6,3.2)}\).
Subtopic 7
Two transformations may be applied successively: the first image becomes the second transformation’s object. The order matters, especially if the transformations involve reflections or rotations.
The textbook uses composition notation: \(PQ\) means apply Q first, then P, just as in composite functions. To avoid mistakes, follow the order stated in words, showing the intermediate image when necessary.
Translate by \(\begin{pmatrix}3\-2\end{pmatrix}\), followed by \(\begin{pmatrix}-5\7\end{pmatrix}\). Give the one translation with the same effect.
Rotate \(P(x,y)\) \(90^\circ\) anticlockwise about the origin, then reflect in the y-axis. Describe the equivalent single transformation.
Rotate: \((x,y)\mapsto(-y,x)\). Reflect the result in the \(y\)-axis:
\[(-y,x)\mapsto(y,x)\]Interchanging the two coordinates is \(\boxed{\text{reflection in }y=x}\).
Reflect a shape in \(y=3\), then in \(y=-1\). Describe the equivalent single transformation.
First: \(y\mapsto6-y\). Then: \(y\mapsto-2-(6-y)=y-8\). The horizontal coordinate is unchanged.
Translation by \(\boxed{\begin{pmatrix}0\\-8\end{pmatrix}}\).
First enlarge by factor \(3\) about \((3,3)\), then translate by \(\begin{pmatrix}2\2\end{pmatrix}\). Find the equivalent single enlargement.
The first operation sends \(P\) to \(3P-2(3,3)\). Translating gives \(3P-(4,4)\). For one enlargement of factor 3 and centre \(C\), the result is \(3P-2C\).
\[-2C=(-4,-4)\Longrightarrow C=\boxed{(2,2)}\]So the overall transformation is an enlargement, scale factor 3, centre \(\boxed{(2,2)}\).
Start with \(A(2,1)\). Reflect in the x-axis, rotate the image \(90^\circ\) anticlockwise about the origin, and then translate by \(\begin{pmatrix}3\-2\end{pmatrix}\). Find the final coordinates.
After reflection: \((2,1)\mapsto(2,-1)\).
After rotation: \((2,-1)\mapsto(1,2)\).
After translation: \((1,2)+(3,-2)=\boxed{(4,0)}\).
Do not change the given order.
Begin with the foundational questions before tackling the longer coordinate and reasoning problems. The exam-style questions are original practice questions, not copied past papers. Click Show Answer to reveal a complete worked solution.
A triangle labelled ABC is translated to A′B′C′. Which is the object? What do primes on the labels mean?
The object is triangle ABC; A′B′C′ is its image. A′ is the new position of the original point A, and similarly for B and C.
State whether (i) a rotation and (ii) an enlargement with scale factor 3 preserves size.
(i) Rotation preserves size, so the image is congruent to the object.
(ii) Factor 3 multiplies lengths by 3, so size changes. The image is similar, not congruent.
Reflect \(A(-4,3)\) in the x-axis.
Reflect \(B(5,-2)\) in the y-axis.
Rotate \(C(3,2)\) by 90° anticlockwise about the origin.
Rotate \(D(-3,7)\) through 180° about the origin.
Translate \(P(4,-1)\) by the vector \(\begin{pmatrix}-6\\5\end{pmatrix}\).
Point \(Q(-1,2)\) is translated to \(Q'(3,8)\). State the vector.
Enlarge the point \(R(3,4)\) by factor 2 about the origin.
Enlarge \(S(8,6)\) by scale factor \(\tfrac12\) about the origin.
Enlarge \(T(2,-5)\) by scale factor −1 about the origin.
The image point is the same distance on the opposite side of the centre.
An enlargement has scale factor \(-3\). Give the length scale factor, area scale factor and effect of the minus sign.
Length scale factor \(=|-3|=\boxed{3}\). Area scale factor \(=(-3)^2=\boxed{9}\). The negative sign places the image on the opposite side of the centre.
Reflect \(A(4,-3)\) in the line \(y=2\).
Distance from the point up to \(y=2\) is 5 units, so image is 5 units above the line:
\[y'=2(2)-(-3)=7\] \[\boxed{A'=(4,7)}\]Reflect the vertices \(A(1,4), B(3,2), C(2,5)\) in \(y=-x\).
Use \((x,y)\mapsto(-y,-x)\) on each point.
\[\boxed{A'=(-4,-1),\ B'=(-2,-3),\ C'=(-5,-2)}\]Rotate \(P(5,4)\) through 180° about \((2,1)\).
Rotate triangle with \(A(1,2),B(2,4),C(4,2)\) 90° clockwise about the origin. Give the image vertices.
Each point follows \((x,y)\mapsto(y,-x)\).
\[\boxed{A'=(2,-1),\ B'=(4,-2),\ C'=(2,-4)}\]Triangle ABC has vertices \(A(-2,1), B(1,1), C(0,4)\). Translate by \(\begin{pmatrix}5\\-3\end{pmatrix}\).
Add 5 to each x-coordinate and subtract 3 from each y-coordinate.
\[\boxed{A'=(3,-2),\ B'=(6,-2),\ C'=(5,1)}\]A point is mapped to \((2,-4)\) by translation vector \(\begin{pmatrix}-3\\6\end{pmatrix}\). Find its original coordinates.
Undo the translation by using the negative of the vector:
\[P=(2,-4)-(-3,6)=\boxed{(5,-10)}\]Enlarge \(A(4,1)\) with centre \((1,-1)\) and scale factor 3.
Enlarge \(B(-4,7)\) by factor \(\tfrac12\) about centre \((2,1)\).
Enlarge \(C(6,4)\) by factor \(-2\) about centre \((2,1)\).
\(A(3,5)\) maps to \(A'(3,-1)\), and \(B(-2,4)\) maps to \(B'(-2,0)\). State the mirror line.
Midpoints have y-coordinate \((5+(-1))/2=2\) and \((4+0)/2=2\).
Reflection in \(\boxed{y=2}\).
\(A(2,1)\mapsto A'(4,2)\) and \(B(4,1)\mapsto B'(8,2)\). Given that the transformation is an enlargement about the origin, determine its scale factor.
The displacement and corresponding edge lengths double.
\[k=\frac{4}{2}=\frac{8}{4}=\boxed{2}\]Translate by \(\begin{pmatrix}-4\\3\end{pmatrix}\), then \(\begin{pmatrix}7\\-6\end{pmatrix}\). Give the single equivalent translation.
Triangle ABC has vertices A(−2,1), B(1,1), C(0,4). Reflect it in \(x=2\). State the image vertices and explain how you know.
A reflection in \(x=2\) gives \(x'=4-x\); y is unchanged.
\[\boxed{A'=(6,1),\ B'=(3,1),\ C'=(4,4)}\]Each pair of corresponding vertices is the same horizontal distance from the vertical line \(x=2\).
Triangle ABC has vertices \(A(2,2),B(4,2),C(2,5)\). Rotate it \(90^\circ\) anticlockwise about \((1,1)\). Find all three image vertices.
Subtract \((1,1)\), rotate \((u,v)\mapsto(-v,u)\), then add \((1,1)\).
\[A':(1,1)\to(-1,1)\to\boxed{(0,2)}\]\[B':(3,1)\to(-1,3)\to\boxed{(0,4)}\]\[C':(1,4)\to(-4,1)\to\boxed{(-3,2)}\]Under a 180° rotation, \(A(-1,4)\) maps to \(A'(7,-2)\). Determine the centre and show your method.
In a half-turn, the centre is the midpoint of a point and its image:
\[C=\left(\frac{-1+7}{2},\frac{4+(-2)}{2}\right)=\boxed{(3,1)}\]Full description: rotation 180° about \((3,1)\).
An enlargement maps \(A(1,2)\) to \(A'(7,-2)\) and \(B(3,2)\) to \(B'(1,-2)\). Find its scale factor and centre.
AB has displacement \((2,0)\); A′B′ has displacement \((-6,0)\).
\[k=\frac{-6}{2}=\boxed{-3}\]Use \(A'=C-3(A-C)=4C-3A\), so \(C=(A'+3A)/4\).
\[C=\frac{(7,-2)+(3,6)}4=\boxed{(2.5,1)}\]An L-shaped polygon has perimeter 30 cm and area 64 cm². Its image is an enlargement of scale factor \(\tfrac34\). Find the new perimeter and area; say if the shapes are congruent.
Lengths and perimeter multiply by \(\tfrac34\):
\[P'=30\times\tfrac34=\boxed{22.5\text{ cm}}\]Area multiplies by \((\tfrac34)^2=\tfrac9{16}\):
\[A'=64\times\tfrac9{16}=\boxed{36\text{ cm}^2}\]They are similar, not congruent, because their sizes differ.
Point \(P(1,0)\) maps to \(P'(5,0)\) and \(Q(2,4)\) maps to \(Q'(4,4)\). Describe the transformation fully.
Midpoints are \((3,0)\) and \((3,4)\), on the vertical line \(x=3\). Segment PP′ and QQ′ are horizontal and perpendicular to it.
\(\boxed{\text{Reflection in }x=3}\).
A shape is reflected in \(y=4\), followed by a reflection in \(y=-2\). Find the equivalent single transformation and the final image of \(P(3,1)\).
The lines are 6 units apart. The resulting translation is 12 units down:
\[\boxed{\begin{pmatrix}0\\-12\end{pmatrix}}\]Check by coordinates: \(y\mapsto8-y\mapsto-4-(8-y)=y-12\).
\[P'=(3,1-12)=\boxed{(3,-11)}\]Apply a 90° clockwise rotation about the origin, followed by a reflection in the x-axis. Describe the equivalent single transformation and find the image of \(P(2,5)\).
Rotate \((x,y)\mapsto(y,-x)\); reflect in x-axis \((y,-x)\mapsto(y,x)\).
Single transformation: \(\boxed{\text{reflection in }y=x}\). So \(P(2,5)\mapsto\boxed{(5,2)}\).
Reflect \(A(-2,3)\) in the x-axis and then in the y-axis. Identify the single equivalent transformation and the final coordinates.
First: \((-2,3)\mapsto(-2,-3)\). Then: \((-2,-3)\mapsto(2,-3)\).
These two perpendicular reflections equal a \(\boxed{180^\circ\text{ rotation about the origin}}\).
Image: \(\boxed{(2,-3)}\).
Triangle ABC has vertices \(A(0,1),B(2,1),C(0,3)\). First translate by \(\begin{pmatrix}3\\-1\end{pmatrix}\), then reflect in the y-axis, then rotate 180° about the origin. Give final coordinates.
After translation: \((3,0),(5,0),(3,2)\).
After reflection in the y-axis: \((-3,0),(-5,0),(-3,2)\).
After half-turn: \(\boxed{A'=(3,0),\ B'=(5,0),\ C'=(3,-2)}\).
Enlarge by scale factor 2 about \((4,1)\), then translate by \(\begin{pmatrix}3\\-2\end{pmatrix}\). Describe the single equivalent enlargement, showing the calculation.
Overall image of P is \(2P-C+v\). This equals an enlargement of scale factor 2 about \(C'=C-v\).
\[C'=(4,1)-(3,-2)=\boxed{(1,3)}\]So it is an enlargement, scale factor \(\boxed{2}\), centre \(\boxed{(1,3)}\).
Under an enlargement of scale factor \(-\tfrac12\) about \((2,4)\), the image of point \(P\) is \(P'=(5,1)\). Find the original point \(P\), showing each step.
Let \(C=(2,4)\). Then \(P'-C=(-\tfrac12)(P-C)\), so \(P-C=-2(P'-C)\).
\[P'-C=(3,-3)\]\[P=(2,4)-2(3,-3)=\boxed{(-4,10)}\]Check: \(C-\tfrac12(-6,6)=(5,1)\).
I distinguish an object from its image and understand invariance, congruence and similarity.
I locate the exact mirror line and count perpendicular distances on both sides.
I give angle, direction and centre; I check which way a quarter-turn moves a point.
I write the horizontal and vertical movements as a signed column vector.
I identify centre and factor; I distinguish fractional and negative factors.
I multiply lengths by |k| but areas by k².
I supply every required detail rather than just the transformation name.
I follow the stated order and use intermediate images to avoid confusion.
I use a sharp pencil and ruler, correctly label every vertex, and keep coordinates in the right order.