Answer
We prove that the sides have equal lengths
and then that the diagonals have equal lengths.
(for details, please see "step-by-step")
Work Step by Step
$A=(-2,9), B=(4,6), C=(1,0)$, and $D=(-5,3)$.
We calculate the distances between the vertices (sides of the quadrilateral)
$d(A, B)=\sqrt{(4-(-2))^{2}+(6-9)^{2}}=\sqrt{6^{2}+(-3)^{2}}$
$=\sqrt{36+9}=\sqrt{45}$
$d(B, C)=\sqrt{(1-4)^{2}+(0-6)^{2}}=\sqrt{(-3)^{2}+(-6)^{2}}$
$=\sqrt{9+36}=\sqrt{45}$
$d(C, D)=\sqrt{(-5-1)^{2}+(3-0)^{2}}=\sqrt{(-6)^{2}+(-3)^{2}}$
$=\sqrt{36+9}=\sqrt{45}$
$d(D, A)=\sqrt{(-2-(-5))^{2}+(9-3)^{2}}=\sqrt{3^{2}+6^{2}}$
$=\sqrt{9+36}=\sqrt{45}$.
So, thus far, we know that ABCD is a rhombus.
$\color{red}{\text{If it is a square, its diagonals will have equal lengths.}}$
$d(A, C)=\sqrt{(1-(-2))^{2}+(0-9)^{2}}=\sqrt{3^{2}+(-9)^{2}}$
$=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$,
$d(B, D)=\sqrt{(-5-4)^{2}+(3-6)^{2}}=\sqrt{(-9)^{2}+(-3)^{2}}$
$=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$.
Since the diagonals are equal, ABCD is a square.
