The vertices of a figure are (-2, 2), (-2, -1 | vedclass

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(C) Let the vertices be $A(-2, 2), B(-2, -1), C(3, -1),$ and $D(3, 2)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x

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Rectangle

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(C) Let the vertices be $A(-2, 2), B(-2, -1), C(3, -1),$ and $D(3, 2)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(-2 - (-2))^2 + (-1 - 2)^2} = \sqrt{0^2 + (-3)^2} = 3$$BC = \sqrt{(3 - (-2))^2 + (-1 - (-1))^2} = \sqrt{5^2 + 0^2} = 5$$CD = \sqrt{(3 - 3)^2 + (2 - (-1))^2} = \sqrt{0^2 + 3^2} = 3$$DA = \sqrt{(-2 - 3)^2 + (2 - 2)^2} = \sqrt{(-5)^2 + 0^2} = 5$Since opposite sides are equal ($AB = CD = 3$ and $BC = DA = 5$) and adjacent sides are perpendicular (as sides are parallel to the coordinate axes),the figure is a rectangle.

The vertices of a figure are $(-2, 2), (-2, -1), (3, -1),$ and $(3, 2)$. This figure is a:

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