Question

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

Answer 1

Let points (-2, -1), (4, 0), (3, 3), and (-3, 2) be respectively denoted by A, B, C, and D.

Slope of $AB = \frac {0+1}{4+2}=\frac 16$

Slope of $CD =\frac {2-3}{-3-3}=\frac {-1}{-6}=\frac 16$

⇒ Slope of AB = Slope of CD

⇒ AB and CD are parallel to each other.

Now, slope of $BC = \frac {3-0}{3-4}=\frac {3}{-1}=-3$

Slope of $AD =\frac {2+1}{-3+2}=\frac {3}{-1}=-3$

⇒ Slope of BC = Slope of AD

⇒ BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.

Thus, points (-2, -1), (4, 0), (3, 3), and (-3, 2) are the vertices of a parallelogram.