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Question
Without using the distance formula, show that points (−2, −1), (4, 0), (3, 3) and (−3, 2) are the vertices of a parallelogram.
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Solution
Let A (−2, −1), B (4, 0), C (3, 3) and D (−3, 2) be the given points.
Now, slope of AB \[= \frac{0 + 1}{4 + 2} = \frac{1}{6}\]
Slope of BC \[= \frac{3 - 0}{3 - 4} = - 3\]
Slope of CD \[= \frac{2 - 3}{- 3 - 3} = \frac{1}{6}\]
Slope of DA \[= \frac{- 1 - 2}{- 2 + 3} = - 3\]
Clearly, we have,
Slope of AB = Slope of CD
Slope of BC = Slope of DA
As the slopes of opposite sides are equal,
Therefore, both pair of opposite sides are parallel.
Hence, the given points are the vertices of a parallelogram.
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