Question
Without using the distance formula, show that 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 a quadrilateral. slope of $\text{AB}=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}$ slope of $\text{BC}=\frac{3-0}{3-4}=-3$ slope of $\text{CD}=\frac{3-2}{3-(-3)}=\frac{1}{6}$ slope of $\text{DA}=\frac{2-(-1)}{-3-(-2)}=-3$ we observe that the slope of opposite side of the quadrilateral ABCD are equal. Hence the quadrilateral ABCD is a parallelogram.

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