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9709 P12 - Nov 2014 - Q9
613
The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90°. The coordinates of A, B, and C are (2, 6), (5, -3), and (8, 3) respectively.
Find the equation of AD.
Find, by calculation, the coordinates of D.
The point E is such that ABCE is a parallelogram.
Find the length of BE.
Solution
(i) To find the equation of AD, we first find the slope of AB which is \(m_{AB} = \frac{-3 - 6}{5 - 2} = -3\). Since AD is perpendicular to AB, the slope of AD is \(m_{AD} = \frac{1}{3}\). Using point A (2, 6), the equation of AD is \(y - 6 = \frac{1}{3}(x - 2)\) or \(3y = x + 16\).
(ii) To find the coordinates of D, we use the equation of CD: \(y - 3 = -3(x - 8)\) or \(y = -3x + 27\). Solving the equations of AD and CD simultaneously, we find \(D = (6\frac{1}{2}, 7\frac{1}{2})\).
(iii) To find the length of BE, we use the midpoint formula to find E such that ABCE is a parallelogram. The midpoint of AC is \((5, 4.5)\), so E is \((5, 12)\). The length of BE is \(\sqrt{(5 - 5)^2 + (12 + 3)^2} = 15\).
