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

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Solution

$(- 2, - 1), (4, 0), (3, 3) a n d (- 3, 2) a r e v e r t i c e s o f p a r a l l e l o g r a m I n p a r a l l e l o g r a m d i a g o n a l s a r e i n t e r s e c t a t s a m e p o i n t s, ∴ m i d p o i n t s d i a g o n a l A C \Rightarrow A C = \frac{3 - 2}{2}, \frac{3 - 1}{2} = (\frac{1}{2}, 1) a n d m i d p o i n t o f d i a g o n a l 3 D i s \Rightarrow B D = (\frac{4 - 3}{2}), \frac{2 + 0}{2} = (\frac{1}{2}, 1)$

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