Every Transformations question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
draw the triangle with vertices \((3,6),(5,6),(3,9)\).
draw the enlarged shape with vertices \((4,6),(8,6),(8,3),(6,3)\).
Triangle \(A\) has vertices \((3,4),(4,4),(3,6)\); the corresponding vertices of \(B\) are \((5,5),(7,5),(5,9)\). The lengths have doubled.
Read corresponding vertices and compare side lengths.
For enlargement factor \(2\), the centre \(O\) satisfies \(B=O+2(A-O)\), so \(O=2A-B\). Using \((3,4)\mapsto(5,5)\) gives \(O=(6,8)-(5,5)=(1,3)\).
Find the fixed centre from a corresponding pair; the other pairs give the same centre.
Add \((4,-3)\) to every vertex: \((3,4)\mapsto(7,1)\), \((4,4)\mapsto(8,1)\), and \((3,6)\mapsto(7,3)\). Plot these three points, join them with straight sides and label the triangle \(C\).
A translation moves every point by the same vector.
(a) Enlargement with scale factor \(2\) and centre \((1,3)\).
(b) \(C:(7,1),(8,1),(7,3)\).
(a) Reflection in \(y=x\) swaps each coordinate:
The image of \(A\) has vertices
(b) Enlarge \(B\) by scale factor \(2\) from centre \((1,1)\).
(a) vertices \((2,5),(2,8),(3,7),(3,6)\); (b) vertices \((5,3),(9,3),(7,7),(5,7)\).
\(a\) Rotate triangle A 90° anticlockwise about (1,2); label triangle B. (b) Describe fully the transformation mapping C onto D.
The rule for 90° anticlockwise rotation about (*a*,*b*): (*x*, *y*) → (*a* − (*y*−*b*), *b* + (*x*−*a*)) About (1,2): (*x*,*y*) → (3−*y*, *x*+1)
Triangle A has vertices *A*<sub>1</sub>(2,7), *A*<sub>2</sub>(4,5), *A*<sub>3</sub>(2,5). Apply the rule: (2,7) → (−4,3), (4,5) → (−2,5), (2,5) → (−2,3)
Triangle B has vertices (−4,3), (−2,5), (−2,3)
Identify corresponding vertices of C and D from the diagram and compute the vector: $$(4,\\,4) \\longrightarrow (-5,\\,7): \\quad \\vec{v} = \\begin{pmatrix}-9\\\\3\\end{pmatrix}$$ Check: (6,1) → (−3,4) gives the same vector $\\begin{pmatrix}-9\\\\3\\end{pmatrix}$
Translation by vector $\\begin{pmatrix}-9\\\\3\\end{pmatrix}$
(a) Shape \(A\) moves to shape \(B\) by the vector
So the transformation is a translation by \(\begin{pmatrix}3\\-5\end{pmatrix}\).
(b) For a \(180^\circ\) rotation about \((-1,0)\),
(a) translation by \(\begin{pmatrix}3\\-5\end{pmatrix}\); (b) vertices \((-4,-1),(-6,-1),(-5,-2),(-4,-2)\).
(a) Triangle \(A\) is enlarged to triangle \(B\).
Using corresponding vertices gives centre \((3,0)\).
So the transformation is an enlargement, scale factor \(3\), centre \((3,0)\).
(b) Translating triangle \(A\) by
(a) enlargement, scale factor \(3\), centre \((3,0)\); (b) vertices \((1,4),(2,4),(1,6)\).
(a) Enlarge from centre \((0,1)\) with scale factor \(3\).
(b) Triangle \(P\) is mapped to triangle \(Q\) by a \(180^\circ\) rotation about \((5,5)\).
(a) vertices \((3,1),(9,1),(9,10),(6,10),(6,4),(3,4)\); (b) rotation \(180^\circ\) about \((5,5)\).
(a) Reflection in \(y=x\) swaps the coordinates:
The reflected T-shape has vertices
(b) Triangle \(A\) is mapped to \(B\) by an enlargement with scale factor \(3\) and centre \((7,2)\).
(a) reflected vertices \((6,1),(6,4),(5,4),(5,3),(4,3),(4,2),(5,2),(5,1)\); (b) enlargement, scale factor \(3\), centre \((7,2)\).
(a) For a \(180^\circ\) rotation about \((-3,2)\),
(c) Triangle \(T\) is mapped to triangle \(U\) by an enlargement with scale factor \(2\) and centre \((-3,3)\).
(a) vertices \((-3,9),(-9,9),(-9,6),(-6,6)\); (b) vertices \((7,3),(13,3),(13,6),(10,6)\); (c) enlargement, scale factor \(2\), centre \((-3,3)\).
(a) The translation from \((-7,2)\) to \((6,12)\) is
So
and
Therefore
(b) Shape \(P\) maps to \(Q\) by an enlargement with scale factor \(3\) and centre \((0,2)\).
(c) Rotate \(P\) \(90^\circ\) clockwise about \((3,5)\). The image has vertices
(a) \(d=5,\ e=7,\ f=-4\); (b) enlargement, scale factor \(3\), centre \((0,2)\); (c) vertices \((3,6),(1,6),(0,5),(1,5)\).
Elite Mathematics · Questions classified by topic, solutions checked by Dr Eslam
Elite Mathematics