The vertices of a quadrilateral ABCD are A (6, -2) , B (9, 2) , C (5, -1) and D (2, -5) . Prove that ABCD is a…

$$\begin{aligned}& AB = \sqrt{(9- 6)^2 + (2 + 2)^2} = 5 \\ & BC = \sqrt{(9-5)^2 + (2 + 1)^2} = 5 \\ & CD = \sqrt{(5-2)^2 + (-1 + 5)^2} = 5 \\ & AD = \sqrt{(6-2)^2 + (-2 + 5)^2} = 5 \\ & AC = \sqrt{(6-5)^2 + (-2 + 1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} \\ & BD = \sqrt{(9-2)^2 + (2 + 5)^2} = \sqrt{7^2 + 7^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2} \\ & As AB = BC = CD = DA\end{aligned}$$ and $$\begin{aligned}& AC \neq BD \\ & \therefore\end{aligned}$$ ABCD is a rhombus and not a square.