2026-08-31T11:34:14.368081 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL The figure shows the quadrilateral \(ABCD\) with ...

Eight marks of coordinate geometry on the quadrilateral \(ABCD\) with vertices \(A(1, 2)\), \(B(5, 5)\), \(C(10, 5)\) and \(D(6, 2)\), whose diagonals \(AC\) and \(BD\) are shown as broken lines in the figure. Three tools cover the whole question: the distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), the gradient formula \(\dfrac{y_2 - y_1}{x_2 - x_1}\), and the property that perpendicular gradients multiply to \(-1\).

(a) Show that \(ABCD\) is a rhombus [3 marks]

A rhombus is a quadrilateral with all four sides equal, so the definition tells you exactly what to compute. Take the sides in order around the shape: \[AB = \sqrt{(5 - 1)^2 + (5 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\] \[BC = \sqrt{(10 - 5)^2 + (5 - 5)^2} = \sqrt{25 + 0} = 5 \quad \textbf{M1 A1}\] \[CD = \sqrt{(6 - 10)^2 + (2 - 5)^2} = \sqrt{16 + 9} = 5\] \[DA = \sqrt{(1 - 6)^2 + (2 - 2)^2} = \sqrt{25} = 5 \quad \textbf{A1}\] \[\text{All four sides equal 5, so } ABCD \text{ is a rhombus.} \quad \textbf{A1 (conclusion)}\]

The three marks are best read as: correct use of the distance formula on the sides (M1), all four lengths correct (A1), and the conclusion drawn (A1). Because this is a "show that", the closing sentence is worth a mark in its own right. Note that computing only two adjacent sides is not enough: a kite can have two pairs of equal adjacent sides without being a rhombus, so all four must be checked.

(b) An equation of the diagonal \(AC\) [2 marks]

\[\text{Gradient of } AC = \frac{5 - 2}{10 - 1} = \frac{3}{9} = \frac{1}{3} \quad \textbf{M1}\] \[y - 2 = \tfrac{1}{3}(x - 1) \ \Rightarrow\ 3y - 6 = x - 1 \ \Rightarrow\ x - 3y + 5 = 0 \quad \textbf{A1}\]

The A1 requires the demanded form \(ax + by + c = 0\) with integer coefficients, so the fraction must be cleared. Using \(C\) rather than \(A\) as the fixed point gives the same equation: \(y - 5 = \tfrac{1}{3}(x - 10)\) rearranges to \(x - 3y + 5 = 0\) as well.

(c) Show \(AC\) and \(BD\) are perpendicular, and find their meeting point [3 marks]

\[\text{Gradient of } BD = \frac{2 - 5}{6 - 5} = \frac{-3}{1} = -3 \quad \textbf{M1}\] \[\tfrac{1}{3} \times (-3) = -1, \ \text{so } AC \perp BD \quad \textbf{A1}\]

The product being \(-1\) is the criterion, and stating that product explicitly is what earns the A1; simply writing the two gradients is not a proof.

For the meeting point, the diagonals of a rhombus bisect each other, so the intersection is the common midpoint. The midpoint of \(AC\) is \[\left(\frac{1 + 10}{2},\ \frac{2 + 5}{2}\right) = \left(\frac{11}{2},\, \frac{7}{2}\right),\] and the midpoint of \(BD\) is \(\left(\dfrac{5 + 6}{2},\ \dfrac{5 + 2}{2}\right) = \left(\dfrac{11}{2},\, \dfrac{7}{2}\right)\), the same point. \(\quad \textbf{A1}\)

"Write down" signals that no solving is expected: the rhombus property does the work. Solving \(x - 3y + 5 = 0\) simultaneously with the equation of \(BD\) is also acceptable and reaches the same point, but it is longer.

Common wrong turns on this question

• Checking only that opposite sides are parallel in part (a). That proves a parallelogram, not a rhombus; equal side lengths are what is required.
• Computing the diagonals \(AC\) and \(BD\) instead of the sides. Here \(AC \ne BD\), and equal diagonals would in fact indicate a rectangle, not a rhombus.
• Sign errors in the gradient of \(BD\), from subtracting the coordinates in a different order in numerator and denominator. Fix an order and keep it.
• Leaving the equation as \(y = \tfrac{1}{3}x + \tfrac{5}{3}\) when integers are demanded.
• Averaging all four vertices to find the intersection. That works for a parallelogram by accident, but the reliable statement is that each diagonal is bisected.

How to check

Confirm the intersection lies on \(AC\): substituting into \(x - 3y + 5 = 0\) gives \(\tfrac{11}{2} - \tfrac{21}{2} + 5 = -5 + 5 = 0\). Confirm the shape another way: \(\overrightarrow{AB} = (4, 3)\) and \(\overrightarrow{DC} = (4, 3)\) are equal, so \(ABCD\) is a parallelogram, and with all sides 5 it is a rhombus. The perpendicular diagonals are then a consequence, which is a useful consistency check on part (c).