Question

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

Answer 1

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.