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Q. If the area of the quadrilateral whose angular points taken in order are (1, 2), (-5, 6), (7, -6), (h, -2) be zero. Show that h = 3.

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Solution

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$G i v e n : A (1, 2), B (- 5, 6), C (7, - 6) a n d D (h, - 2) W e k n o w t h a t A r e a o f q u a d r i l a t e r a l = A r e a o f △ A B C + A r e a o f △ A C D a n d A r e a o f △ A B C = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| N o w, A r e a o f △ A B C = \frac{1}{2} |1 (6 + 6) - 5 (- 6 - 2) + 7 (2 - 6)| = \frac{1}{2} |12 + 40 - 28| = 12 A r e a o f △ A C D = \frac{1}{2} |- 5 (- 6 + 2) + 7 (- 2 - 6) + h (6 + 6)| = \frac{1}{2} |20 - 56 + 12 h| = 6 h - 18 G i v e n a r e a o f q u a d r i l a t e r a l = 0 S o, 12 + 6 h - 18 = 0 \Rightarrow 6 h - 6 = 0 \Rightarrow 6 h = 6 \Rightarrow h = 1$
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