The children's corner of a community library is planned on squared paper. A soft play mat is drawn as triangle \(Q\), with vertices \((1,1)\), \((4,1)\) and...

Triangle \(Q\) has vertices (1, 1), (4, 1) and (1, 3). Both transformations here have a centre or mirror line away from the origin, so the rules must be built from the geometry rather than quoted: a half turn about \((a, b)\) sends \((x, y)\) to \((2a - x, \ 2b - y)\), and a reflection in \(y = b\) sends \((x, y)\) to \((x, \ 2b - y)\).

1. Rotation of 180 degrees about (3, 2) [3 marks]
   The centre is (3, 2), so the rule is \[(x, y) \rightarrow (6 - x, \ 4 - y) \tag{1 mark}\] \[(1, 1) \rightarrow (5, 3), \qquad (4, 1) \rightarrow (2, 3) \tag{1 mark}\] \[(1, 3) \rightarrow (5, 1) \tag{1 mark}\] The image has vertices (5, 3), (2, 3) and (5, 1). Each image point is the same distance from (3, 2) as its original but on the opposite side, which is a useful check: (1, 1) is 2 left and 1 below the centre, and (5, 3) is 2 right and 1 above it.
2. Reflection in the line \(y = 2\) [2 marks]
   The mirror is horizontal, so \(x\) is unchanged and \(y\) maps to \(4 - y\): \[(1, 1) \rightarrow (1, 3), \qquad (4, 1) \rightarrow (4, 3) \tag{1 mark}\] \[(1, 3) \rightarrow (1, 1) \tag{1 mark}\] The second image has vertices (1, 3), (4, 3) and (1, 1).
3. Transformation from the second image to the first [2 marks]
   Match the vertices: (1, 3) goes to (5, 3), (4, 3) goes to (2, 3) and (1, 1) goes to (5, 1). The \(y\) coordinates are unchanged and each \(x\) becomes \(6 - x\), so the mirror line is \(x = 3\). The single transformation is a reflection in the line \(x = 3\). This is consistent with the earlier parts, since a reflection in \(y = 2\) followed by a reflection in \(x = 3\) gives the half turn about their crossing point (3, 2).
4. Why all three triangles have equal areas [2 marks]
   A rotation and a reflection are both isometries: they preserve every length and every angle. Each image is therefore congruent to \(Q\), and congruent shapes have equal areas. Only an enlargement with scale factor other than \(\pm 1\) would change the area. Here that area is \(\tfrac{1}{2} \times 3 \times 2 = 3\) square units for all three triangles.

Examination reminder: two reflections in perpendicular lines always combine to give a half turn about the point where the lines cross, which is a quick way to check answers like these against each other.