Show that the points $A(2, 2)$,$B(-2, 2)$,$C(-2, -2)$,and $D(2, -2)$ are the vertices of a square.

(A) Let the given points be $A(2, 2)$,$B(-2, 2)$,$C(-2, -2)$,and $D(2, -2)$.
To prove that these points form a square,we must show that all four sides are equal and both diagonals are equal.
$1$. Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-2 - 2)^2 + (2 - 2)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.
$BC = \sqrt{(-2 - (-2))^2 + (-2 - 2)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$ units.
$CD = \sqrt{(2 - (-2))^2 + (-2 - (-2))^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$ units.
$DA = \sqrt{(2 - 2)^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.
Since $AB = BC = CD = DA = 4$,all sides are equal.
$2$. Calculate the lengths of the diagonals:
$AC = \sqrt{(-2 - 2)^2 + (-2 - 2)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
$BD = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ units.
Since the sides are equal and the diagonals are equal,the points $A, B, C, D$ form a square.