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Question
Show that A(–4, –7), B (–1, 2), C (8, 5) and D (5, –4) are the vertices of a parallelogram.
Solution
The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
Slope of AB = \[\frac{2 - \left( - 7 \right)}{- 1 - \left( - 4 \right)} = \frac{9}{3} = 3\]
Slope of BC = \[\frac{5 - 2}{8 - \left( - 1 \right)} = \frac{3}{9} = \frac{1}{3}\]
Slope of CD = \[\frac{- 4 - 5}{5 - 8} = \frac{- 9}{- 3} = 3\]
Slope of AD = \[\frac{- 4 - \left( - 7 \right)}{5 - \left( - 4 \right)} = \frac{3}{9} = \frac{1}{3}\]
Slope of AB = Slope of CD
Slope of BC = Slope of AD
So, AB || CD and BC || AD
Hence, ABCD is a parallelogram.
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