Question

Quadrilateral ABCD has vertices A(-3,6), B(6,0),C(9,-9), and D(0, -3). Prove that ABCD isa) a parallelogramb) not a rhombus

Answer 1

We have the following vertices given:

A(-3,6), B=(6,0), C(9,-9), D=(0,-3)

a) We can begin find the distance between two points using this formula:

$d=\text{ }\sqrt{(x_2-x_1)^2+(y_{2\text{ }}-y_1)^2}$

For this case we need to find the distance between AD and BC and using the formula given we have:

$AD=\text{ }\sqrt{(0+3)^2+(-3-6)^2}=\text{ }\sqrt{90}=\text{ 3}\sqrt{10}$$BC=\text{ }\sqrt{(9-6)^2+(-9-0)^2}=\text{ }\sqrt{90}=\text{ 3}\sqrt{10}$

We see that both distances are equal. Now we can calculate the slopes for AD and BC and we got using this formula:

$m=\text{ }\frac{y_2-y_1}{x_{2\text{ }}-x_1}$$_{}m_{AD}=\text{ }\frac{-3-6}{0+3}=\text{ }-3$$m_{BC}=\frac{-9-0}{9-6}=-3$

Then we can conclude that the quadrilateral is a paralellogram because it has one pair of opposite sides that both congruent and parallel

b) For this problem is not a rhombus because not all the distances AB, AC, AD, BC and CD are equal