Transformations – Year 8 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Reflecting in the x-axis keeps each x-coordinate and changes the sign of the y-coordinate: \((x,y)\to(x,-y)\). The image is \((1,-2)\), \((4,-2)\), \((1,-5)\). (1 mark for two correct, 2 marks for all three)
   2. The vector \(\begin{pmatrix}-3\\2\end{pmatrix}\) means 3 units left and 2 units up, so subtract 3 from each x and add 2 to each y. The image is \((-2,4)\), \((1,4)\), \((-2,7)\). (1 mark for two correct, 2 marks for all three)

   Both transformations keep the triangle the same size and shape, so the images are congruent to ABC. Check (a): each image point is the same distance below the axis as the original is above it. Check (b): AB is still 3 units long and horizontal.

2. Answer to Question 2 (4 marks)

   1. A rotation of 90° anticlockwise about the origin sends \((x,y)\) to \((-y,x)\). So \((2,1)\to(-1,2)\), \((4,1)\to(-1,4)\) and \((2,3)\to(-3,2)\). The image is \((-1,2)\), \((-1,4)\), \((-3,2)\). (1 mark for two correct, 2 marks for all three)
   2. Every coordinate has changed sign: \((x,y)\to(-x,-y)\). This is a rotation of 180° about the origin (1 mark for rotation, 1 mark for both the angle and the centre).

   Tracing paper is a good check for (a): trace the triangle, pin the origin and turn the paper a quarter turn anticlockwise. For a 180° rotation the direction does not need to be stated, because clockwise and anticlockwise give the same image.

3. Answer to Question 3 (4 marks)

   1. Multiply each length by the scale factor 4: \(5\times4=20\) and \(3\times4=12\). The image is 20 cm by 12 cm. (1 mark)
   2. Perimeter of R \(=2(5+3)=\)16 cm; image \(=2(20+12)=\)64 cm. The perimeter is multiplied by 4, the same as the lengths. (1 mark)
   3. Area of R \(=5\times3=\)15 cm² (1 mark). Image \(=20\times12=\)240 cm² (1 mark).

   The area has been multiplied by \(240\div15=16\), which is \(4^{2}\). Lengths grow by the scale factor, but areas grow by its square, because both the length and the width are multiplied.

4. Answer to Question 4 (4 marks)

   1. Scale factor = new length ÷ old length \(=60\div8=\)7.5. (1 mark)
   2. Areas are multiplied by the square of the scale factor: \(7.5^{2}=56.25\) (1 mark). The banner logo has area \(30\times56.25=\)1 687.5 cm² (1 mark). There are \(100\times100=10\,000\) cm² in 1 m², so this is \(1687.5\div10000=0.16875\), about 0.17 m² (1 mark).

   Multiplying 30 by 7.5 gives only 225 cm², a common mistake that forgets areas scale twice. Dividing by 100 instead of 10 000 in the last step is the other error to avoid.

5. Answer to Question 5 (4 marks)

   1. The mirror line \(x=3\) is vertical, so the y-coordinates stay the same. Each point ends the same distance on the other side of the line: \((x,y)\to(6-x,y)\). For example, \((1,1)\) is 2 units left of the line, so its image is 2 units right, at \((5,1)\). The vertices of \(T_1\) are \((5,1)\), \((4,1)\), \((5,3)\) (1 mark for two, 2 marks for all three).
   2. An enlargement by scale factor 2 from the origin doubles every coordinate. The vertices of \(T_2\) are \((10,2)\), \((8,2)\), \((10,6)\) (1 mark for two, 2 marks for all three).

   Check: the sides of \(T_2\) are twice as long as those of \(T_1\).

Total : 20 marks