Question

Without using distance formula, show that the points A(1, –2), B(3, 6), C(5, 10) and D(3, 2) are the vertices of a parallelogram.

Answer 1

A quadrilateral is a parallelogram if both pairs of opposite sides are parallel (i.e. have equal slopes).

Slope of side $$\mathrm{AB}$$:

$$m_{\mathrm{AB}} = \frac{6 - (-2)}{3 - 1}$$

$$= \frac{8}{2}$$ 

$$= 4$$

Slope of side $$\mathrm{CD}$$: 

$$m_{\mathrm{CD}} = \frac{2 - 10}{3 - 5}$$

$$= \frac{-8}{-2}$$

$$= 4$$

Since $$m_{\mathrm{AB}} = m_{\mathrm{CD}}$$, $$\mathrm{AB} \parallel \mathrm{CD}$$.

Slope of side $$\mathrm{BC}$$: 

$$m_{\mathrm{BC}} = \frac{10 - 6}{5 - 3}$$

$$= \frac{4}{2}$$

$$= 2$$

Slope of side $$\mathrm{DA}$$: 

$$m_{\mathrm{DA}} = \frac{-2 - 2}{1 - 3}$$

$$= \frac{-4}{-2}$$

$$= 2$$

Since $$m_{\mathrm{BC}} = m_{\mathrm{DA}}$$, $$\mathrm{BC} \parallel \mathrm{DA}$$.

Because both pairs of opposite sides are parallel, $$\mathrm{ABCD}$$ is a parallelogram.