The points \(A(1,2)\), \(B(7,2)\), \(C(7,8)\) and \(D(1,8)\) are the vertices of a square. (a) Write down the equations of the two lines of symmetry of the ...

First confirm the shape. From \(A(1,2)\) to \(B(7,2)\) is 6 units across, and from \(B(7,2)\) to \(C(7,8)\) is 6 units up, so \(ABCD\) is a square of side 6 with centre at \((4,5)\).

(a) Lines of symmetry parallel to the axes. Each passes through the centre and through the midpoints of a pair of opposite sides.

• Vertical sides lie at \(x=1\) and \(x=7\), so the vertical mirror is \( x = \frac{1+7}{2} = 4 \), giving \( x=4 \) [B1].
• Horizontal sides lie at \(y=2\) and \(y=8\), so the horizontal mirror is \( y = \frac{2+8}{2} = 5 \), giving \( y=5 \) [B1].

Equivalent forms are accepted.

(b) Order of rotational symmetry. All four sides are equal and all four angles are \(90^\circ\), so a quarter turn about \((4,5)\) sends each side onto the next. Matches occur at \(90^\circ\), \(180^\circ\), \(270^\circ\) and \(360^\circ\), so the order is 4 [B1].

(c) The diagonal line of symmetry with positive gradient. A square also has both diagonals as mirror lines. The diagonal with positive gradient runs from \(A(1,2)\) up to \(C(7,8)\).

Gradient 1 seen, or the line through \((1,2)\) and \((7,8)\) identified [M1]

Substituting \((1,2)\) into \(y = x + c\) gives \(2 = 1 + c\), so \(c = 1\).

Equation: \( y = x + 1 \) [A1] (or any equivalent form)

Check it against the centre: \((4,5)\) satisfies \(y=x+1\), as every line of symmetry of the square must pass through the centre. The other diagonal, through \((1,8)\) and \((7,2)\), has gradient \(-1\) and is the one excluded by the words "positive gradient".