Question

quadrilateral DEFG has vertices at D(3,4), E(8,6), F(9,9), and G(4,7). Prove that defg is a parallelogram.

Answer 1

I'll graph it

From the graph we can observe that it is a parallelogram, let's calculate the distance between the vertex to know if it is a parallelogram.

`\begin{gathered} \text{distance DE = }\sqrt[]{(8-3)^2+(6-4)^2} \\ \text{ = }\sqrt[]{(5)^2+(2)^2} \\ \text{ = }\sqrt[]{25\text{ + 4}} \\ \text{ = }\sqrt[]{29} \\ dis\tan ce\text{ DG = }\sqrt[]{(4-3)^2+(7-4)^2} \\ \text{ = }\sqrt[]{(1)^2+(3)^2} \\ \text{ = }\sqrt[]{1\text{ + 9}} \\ \text{ = }\sqrt[]{10} \\ \text{distance EF = }\sqrt[]{(9-8)^2+(9-6)^2} \\ \text{ = }\sqrt[]{(1)^2+(3)^2} \\ \text{ = }\sqrt[]{1\text{ + 9}} \\ \text{ = }\sqrt[]{10} \\ \text{distance GF = }\sqrt[]{(9-4)^2+(9-7)^2} \\ \text{ = }\sqrt[]{(5)^2+(2)^2} \\ \text{ = }\sqrt[]{25\text{ + 4}} \\ \text{ = }\sqrt[]{29} \end{gathered}`

Conclusion: it is a parallelogram because DE = GF

and DG = EF