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Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

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Let the vertices be A(-2, -1), B(4, 0), C(3, 3), D(-3, 2)

Now mAB = mCD ⇒ AB || CD

and mBC = mAD ⇒ BC || AD

Hence ABCD is a parallelogram.

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