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

Answer

Let A(-2, -1), B (4, 0), C (3, 3) and D(-3, 2) be vertices of a quadrilateral ABCD.
$\therefore$ Slope of AB $ = \frac{{0 - ( - 1)}}{{4 - ( - 2)}} = \frac{1}{6}$
Slope of BC $= \frac{{3 - 0}}{{3 - 4}} = \frac{3}{{ - 1}} = - 3$
Slope of DC $= \frac{{3 - 2}}{{3 - ( - 3)}} = \frac{1}{6}$
Slope of AD $= \frac{{2 - ( - 1)}}{{ - 3 - ( - 2)}} = \frac{3}{{ - 1}} = - 3$
$\therefore$ Slope of AB = Slope of DC $\Rightarrow$ AB|| DC
And Slope of BC = Slope of AD $\Rightarrow$ BC || AD
Thus ABCD is a parallelogram

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