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Sagot :
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:
[tex]d=\text{ }\sqrt{(x_2-x_1)^2+(y_{2\text{ }}-y_1)^2}[/tex]For this case we need to find the distance between AD and BC and using the formula given we have:
[tex]AD=\text{ }\sqrt{(0+3)^2+(-3-6)^2}=\text{ }\sqrt{90}=\text{ 3}\sqrt{10}[/tex][tex]BC=\text{ }\sqrt{(9-6)^2+(-9-0)^2}=\text{ }\sqrt{90}=\text{ 3}\sqrt{10}[/tex]We see that both distances are equal. Now we can calculate the slopes for AD and BC and we got using this formula:
[tex]m=\text{ }\frac{y_2-y_1}{x_{2\text{ }}-x_1}[/tex][tex]_{}m_{AD}=\text{ }\frac{-3-6}{0+3}=\text{ }-3[/tex][tex]m_{BC}=\frac{-9-0}{9-6}=-3[/tex]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
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