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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.

Text Solution

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Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be vertices of a quadrilateral ABCD.
Then, slope of `AB = (0-(-1))/(4-(-2)) = 1/6`
Slope of `BC = (3-0)/(3-4) = -3`
Slope of `CD = (2-3)/(3-(-3)) = -1/6`
Slope of `DA = (2-1)/(-3-(-2)) = -3`
As, slope of `AB` = slope of `CD`
and slope of `BC` = slope of `AD`
...
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