Only one quarter of a bus network symbol, quadrilateral \(EFGH\), is drawn on the grid below. Lines \(x = 6\) and \(y = 6\) are both set to become lines of ...

Reflecting a point in a vertical line \(x = k\) uses \((x, y) \to (2k - x, y)\); reflecting in a horizontal line \(y = k\) uses \((x, y) \to (x, 2k - y)\).

(a) Reading \(E\) as \((2, 2)\), reflecting in \(x = 6\) gives \(x' = 2(6) - 2 = 10\), so the image is \((10, 2)\). [1 mark]

(b) Reflecting \(E(2, 2)\) in \(y = 6\) gives \(y' = 2(6) - 2 = 10\), so the image is \((2, 10)\). [1 mark]

(c) Reflecting \(E\) in \(x = 6\) first gives \((10, 2)\) from part (a); reflecting this result in \(y = 6\) gives \(y' = 2(6) - 2 = 10\), so the final image is \((10, 10)\). [2 marks]

(d) Reflecting in two perpendicular lines in succession, as in part (c), is equivalent to a single \(180^\circ\) rotation about the point where the two lines cross, \((6, 6)\). This gives the finished symbol rotational symmetry of order \(2\). [1 mark]

(e) With \(3\) equally spaced lines of symmetry through a point, a pattern like this has rotational symmetry of order \(3\), since \(n\) equally spaced mirror lines through a point always give rotational symmetry of order \(n\). [1 mark]

(f) The designer's claim is not correct: \(n\) equally spaced lines give rotational symmetry of order \(n\), so \(3\) equally spaced lines give order \(3\), and a third line added to the original \(2\) still only gives order \(3\), not \(4\). [1 mark]

The general rule linking equally spaced mirror lines to rotational order, order equals the number of lines, runs through parts (d) to (f) and is worth learning as a single fact rather than working out each case from scratch.