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{ }√((0+3)^2+(-3-6)^2)=\text{ }√(90)=\text{ 3}√(10)`
`BC=\text{ }√((9-6)^2+(-9-0)^2)=\text{ }√(90)=\text{ 3}√(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{ }(-3-6)/(0+3)=\text{ }-3`
`m_(BC)=(-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