Question

Show that A(–4, –7), B (–1, 2), C (8, 5) and D (5, –4) are the vertices of a parallelogram.

Answer 1

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
Slope of AB = $\frac{2-\left(-7\right)}{-1-\left(-4\right)}=\frac{9}{3}=3$
Slope of BC = $\frac{5-2}{8-\left(-1\right)}=\frac{3}{9}=\frac{1}{3}$
Slope of CD = $\frac{-4-5}{5-8}=\frac{-9}{-3}=3$
Slope of AD = $\frac{-4-\left(-7\right)}{5-\left(-4\right)}=\frac{3}{9}=\frac{1}{3}$
Slope of AB = Slope of CD
Slope of BC = Slope of AD
So, AB || CD and BC || AD
Hence, ABCD is a parallelogram.