Question

Using slopes, show that the vertices (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Answer 1

Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the vertices of the given quadrilateral ABCD.

Then,

Slope of AB = $\frac{0-(-1)}{4-(-2)}=\frac{1}{6}$

Slope of DC = $\frac{3-2}{3-(-3)}=\frac{1}{6}$

Slope of BC = $\frac{3-0}{3-4}=\frac{3}{-1}=-3$

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

$\therefore$ slope of AB = slope of DC $\Rightarrow$ AB || DC;

slope of BC = slope of AD $\Rightarrow$ BC || AD;

Hence, ABCD is a parallelogram.