Year 11 transformations worksheet answers to download as PDF

These Year 11 transformations worksheet answers give a full worked solution to all 21 questions of the transformations worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. Change the sign of each \(x\)-coordinate: \((-1,1)\), \((-4,1)\), \((-1,3)\).
2. Swap \(x\) and \(y\): \((1,1)\), \((1,4)\), \((3,1)\).
3. The mirror is 5 units from the \(y\)-axis; the image of \(x\) is \(10-x\): \((9,1)\), \((6,1)\), \((9,3)\).
4. Reflection in \(y=x\) leaves \(A(1,1)\) unchanged because \(A\) lies on the line.
Answer to Question 2
1. Subtract 3 from each \(x\) and add 4 to each \(y\): \((-2,5)\), \((1,5)\), \((-2,7)\). The top number of the vector moves each point across and the bottom number moves it up or down.
2. The opposite vector: \(\begin{pmatrix}3\\-4\end{pmatrix}\). The reverse translation moves each point back by the same amount in the opposite direction.
3. Vector \(=\begin{pmatrix}6-1\\-2-1\end{pmatrix}=\) \(\begin{pmatrix}5\\-3\end{pmatrix}\). \(B\to(9,-2)\), \(C\to(6,0)\). The vector is the change from \(A\) to its image; apply the same change to \(B\) and \(C\).
Answer to Question 3
1. \((x,y)\to(y,-x)\): \((1,-1)\), \((1,-4)\), \((3,-1)\). A \(90^\circ\) clockwise turn about the origin sends \((x,y)\) to \((y,-x)\). Check: \(B(4,1)\) goes to \((1,-4)\), still 4.12 units from the origin.
2. \((x,y)\to(-y,x)\): \((-1,1)\), \((-1,4)\), \((-3,1)\). A \(90^\circ\) anticlockwise turn sends \((x,y)\) to \((-y,x)\). The triangle moves into the second quadrant.
3. \((x,y)\to(-x,-y)\): \((-1,-1)\), \((-4,-1)\), \((-1,-3)\). A half turn changes the sign of both coordinates, so the triangle ends up in the third quadrant.
Answer to Question 4
1. \(\overrightarrow{CP}=\begin{pmatrix}2\\-1\end{pmatrix}\). Subtract the coordinates of the centre from those of \(P\).
2. \((x,y)\to(-y,x)\) gives \(\begin{pmatrix}1\\2\end{pmatrix}\). Turning a vector \(90^\circ\) anticlockwise sends \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}-y\\x\end{pmatrix}\).
3. Add to the centre: \((2+1,2+2)=\) \((3,4)\). The rotated vector starts at the centre, so we add it to the coordinates of \(C\).
4. \(CP=\sqrt{4+1}=\sqrt5\). \(CP'=\sqrt{1+4}=\sqrt5\). ✓ A rotation keeps every point the same distance from the centre.
Answer to Question 5
1. Multiply every coordinate by 2: \((2,2)\), \((8,2)\), \((2,6)\).
2. Vector from the centre to each point multiplied by 3. \(A\) is at the centre so it stays at \((1,1)\). \(B\): \((1,1)+3(3,0)=\) \((10,1)\). \(C\): \((1,1)+3(0,2)=\) \((1,7)\).
3. Area of \(T=\tfrac12\times3\times2=3\). The image area is \(3\times3^2=\) 27.
Answer to Question 6
1. A reflection in the \(y\)-axis (\(x=0\)). Each \(x\)-coordinate changes sign and each \(y\)-coordinate stays the same.
2. A translation by the vector \(\begin{pmatrix}2\\3\end{pmatrix}\). Every point moves 2 right and 3 up.
3. A rotation of \(180^\circ\) about the origin (centre \((0,0)\)). Both coordinates change sign, which is a half turn about the origin.
Answer to Question 7
1. \((x,y)\to(-y,x)\): \((0,6)\).
2. \((-6,0)\).
3. \(270^\circ\) anticlockwise is \(90^\circ\) clockwise: \((x,y)\to(y,-x)\): \((0,-6)\).
4. Distance from the origin is always 6, so the tip is on a circle of radius 6. One turn is \(2\pi\times6=\) 37.7 cm.
Answer to Question 8
1. Multiply by 1.5: \((3,3)\), \((9,3)\), \((9,7.5)\), \((3,7.5)\). With centre the origin, every coordinate is multiplied by the scale factor.
2. Before: \(4\times3=\) 12 cm². After: \(6\times4.5=\) 27 cm². The original rectangle is \(6-2=4\) cm by \(5-2=3\) cm; the image is \(9-3=6\) cm by \(7.5-3=4.5\) cm.
3. \(27:12=9:4\), which is \(1.5^2:1\), the square of the scale factor. Divide both areas by 3 to simplify the ratio.
Answer to Question 9
1. \(\begin{pmatrix}3-1\\2-1\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}\).
2. Multiply by \(-2\): \(\begin{pmatrix}-4\\-2\end{pmatrix}\). Add to the centre: \((-3,-1)\).
3. The image is on the opposite side of the centre and is turned upside down, and its lengths are 2 times as long.
4. Area scale factor \(=(-2)^2=4\), so the image area is \(5\times4=\) 20 cm².
Answer to Question 10
1. \(\overrightarrow{CS}=\begin{pmatrix}3\\0\end{pmatrix}\). Clockwise: \((x,y)\to(y,-x)\) gives \(\begin{pmatrix}0\\-3\end{pmatrix}\). Add the centre: \((3,0)\).
2. \(\begin{pmatrix}-3\\0\end{pmatrix}\), so \((0,3)\).
3. \(\dfrac{15}{40}\times360=\) \(135^\circ\). 15 seconds is \(\tfrac{15}{40}\) of a full turn.
4. Circumference \(=2\pi\times3=18.85\) cm in 40 s, so speed \(=\) 0.47 cm/s. Speed \(=\) distance \(\div\) time: \(18.85\div40\approx0.47\).
Answer to Question 11
1. The image of \(x\) is \(a+2(x-a)\). With \(x=2\): \(a+2(2-a)=5\). The image of a point is the centre plus 2 times the vector from the centre to the point.
2. \(a+4-2a=5\), so \(-a=1\) and \(a=-1\). By symmetry \(b=-1\). The centre is \((-1,-1)\). The same working with the \(y\)-coordinates, \(b+2(2-b)=5\), gives \(b=-1\).
3. \(\overrightarrow{CQ}=\begin{pmatrix}4\\3\end{pmatrix}\); times 2 gives \(\begin{pmatrix}8\\6\end{pmatrix}\); \(Q'=(-1+8,-1+6)=\) \((7,5)\). The image \(Q'\) is twice as far from the centre as \(Q\), so the centre is correct.
Answer to Question 12
1. After the first: \((-1,1)\), \((-4,1)\), \((-1,3)\). After the second: \((-1,-1)\), \((-4,-1)\), \((-1,-3)\).
2. A rotation of \(180^\circ\) about the origin.
3. Reflecting in the \(x\)-axis first gives \((1,-1)\), then in the \(y\)-axis gives \((-1,-1)\): the same. For these two perpendicular mirrors the order does not matter.
Answer to Question 13
1. Right: \(\begin{pmatrix}2\\0\end{pmatrix}\). Above: \(\begin{pmatrix}0\\2\end{pmatrix}\). Each tile moves 2 units, the side of the tile, to the next position.
2. Translation \(\begin{pmatrix}6\\4\end{pmatrix}\): \((6,4)\), \((8,4)\), \((8,6)\), \((6,6)\). 3 places right is \(3\times2=6\) across and 2 places above is \(2\times2=4\) up.
3. \(6\div0.5=12\) and \(4\div0.5=8\), so \(12\times8=\) 96 tiles. Divide each length of the room by the side of a tile, then multiply.
Answer to Question 14
1. \(8\times2.5=\) 20 cm. Lengths are multiplied by the scale factor.
2. \(12\times2.5^2=12\times6.25=\) 75 cm². Areas are multiplied by the square of the scale factor.
3. \(75\div6.25=\) 12 cm². To go back from the copy to the original, divide by the area factor.
Answer to Question 15
1. First reflection: \((5,2)\). Second: \((-5,2)\). Reflecting in \(y=x\) swaps the coordinates; reflecting in the \(y\)-axis changes the sign of \(x\).
2. \((x,y)\to(-y,x)\) turns \((2,5)\) into \((-5,2)\). So the combination is a rotation of \(90^\circ\) anticlockwise about the origin. The final point matches the rule for a \(90^\circ\) anticlockwise rotation about the origin.
3. \((3,1)\to(1,3)\to(-1,3)\) and \(90^\circ\) anticlockwise gives \((-1,3)\) ✓. The rule works for a second point, which supports the answer.
Answer to Question 16
1. A reflection in the \(x\)-axis (\(y=0\)). The \(x\)-coordinates stay the same and the \(y\)-coordinates change sign.
2. A rotation of \(180^\circ\) about the origin, or an enlargement of scale factor \(-1\) with centre the origin. Both coordinates change sign, so either description gives the same image.
3. \(V\to U\): \((x,y)\to(-x,y)\), a reflection in the \(y\)-axis (\(x=0\)). The \(y\)-coordinates stay the same and the \(x\)-coordinates change sign.
Answer to Question 17
1. \(P'=(1,1)-2(2,1)=\) \((-3,-1)\). \(Q'=(1,1)-2(4,1)=\) \((-7,-1)\). \(R'=(1,1)-2(2,3)=\) \((-3,-5)\). Find the vector from the centre to each point, multiply it by \(-2\), then add it to the centre.
2. Area of \(PQR=\tfrac12\times2\times2=2\). Image \(=2\times(-2)^2=\) 8. The area scale factor is the square of the scale factor, which is positive.
3. A negative scale factor reverses the direction of every vector from the centre.
Answer to Question 18
1. \(20\times3^2=\) 180 cm². Area is multiplied by the square of the scale factor, \(3^2=9\).
2. \(12\times3^3=\) 324 cm³. Volume is multiplied by the cube of the scale factor, \(3^3=27\).
3. Length: \(10\times0.5=\) 5 cm. Area: \(40\times0.25=\) 10 cm². Lengths are multiplied by 0.5 and areas by \(0.5^2=0.25\), so the shape gets smaller.
Answer to Question 19
1. \(A'B'=2\) and \(AB=1\), so the scale factor is 2.
2. \(AA'\): gradient \(\dfrac{4-2}{3-2}=2\), so \(y=2x-2\). \(BB'\): gradient \(\dfrac{4-2}{5-3}=1\), so \(y=x-1\).
3. Solve \(2x-2=x-1\): \(x=1\) and \(y=0\). The centre is \((1,0)\). Check with \(C\): \((1,0)+2(1,4)=(3,8)\) ✓.
Answer to Question 20
1. After reflecting: \((1,-1)\), \((4,-1)\), \((1,-3)\). After translating: \((-1,4)\), \((2,4)\), \((-1,2)\).
2. Yes: reflection and translation both preserve lengths and angles.
3. The combination of a reflection and a translation can be a single reflection only if the translation is perpendicular to the mirror line. Here the vector \((-2,5)\) has a part parallel to the mirror (\(-2\)), so it is a glide reflection, not a single reflection. The midpoint of \(A\) and \(A'\) is \((0,2.5)\), not on the \(x\)-axis.
Answer to Question 21
1. Reflection in \(x=3\): \((5,1)\). Reflection in \(x=7\): \((9,1)\).
2. A translation by \(\begin{pmatrix}8\\0\end{pmatrix}\). The mirrors are 4 apart and \(8=2\times4\).
3. \((2,0)\) stays at \((2,0)\) in the first reflection, then goes to \((0,2)\). This is a rotation of \(90^\circ\) anticlockwise about the origin. The mirrors meet at \(45^\circ\) and \(90^\circ=2\times45^\circ\).

How to use these Year 11 transformations worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the transformations lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 11 transformations worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 11 transformations worksheet answers show every step, so nothing is left unexplained.