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(B) Let the vertices be $A(0, -1), B(2, 1), C(0, 3),$ and $D(-2, 1).$Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y

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Square

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(B) Let the vertices be $A(0, -1), B(2, 1), C(0, 3),$ and $D(-2, 1).$Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(2-0)^2 + (1-(-1))^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{4+4} = 2\sqrt{2}$$CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$$DA = \sqrt{(0-(-2))^2 + (-1-1)^2} = \sqrt{4+4} = 2\sqrt{2}$Calculate the diagonals:$AC = \sqrt{(0-0)^2 + (3-(-1))^2} = \sqrt{0+16} = 4$$BD = \sqrt{(-2-2)^2 + (1-1)^2} = \sqrt{16+0} = 4$Since all sides are equal $(AB=BC=CD=DA)$ and diagonals are equal $(AC=BD)$,the quadrilateral is a square.

If the vertices of a quadrilateral are $(0, -1), (2, 1), (0, 3),$ and $(-2, 1)$,then it is a

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