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(C) The vertices are represented as points in the complex plane: $A(1, 2),$ $B(-3, 1),$ $C(-2, -3),$ and $D(2, -2).$Calculate the lengths of the sides

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Square

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(C) The vertices are represented as points in the complex plane: $A(1, 2),$ $B(-3, 1),$ $C(-2, -3),$ and $D(2, -2).$Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(-3 - 1)^2 + (1 - 2)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$$BC = \sqrt{(-2 - (-3))^2 + (-3 - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$$CD = \sqrt{(2 - (-2))^2 + (-2 - (-3))^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}$$DA = \sqrt{(1 - 2)^2 + (2 - (-2))^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$Since all sides are equal,it is a rhombus. Now,check the diagonals:$AC = \sqrt{(-2 - 1)^2 + (-3 - 2)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$$BD = \sqrt{(2 - (-3))^2 + (-2 - 1)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$Since all sides are equal and the diagonals are equal,the quadrilateral is a square.

If the vertices of a quadrilateral are $A = 1 + 2i,$ $B = -3 + i,$ $C = -2 - 3i,$ and $D = 2 - 2i,$ then the quadrilateral is:

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