What type of a quadrilateral do the points A( | vedclass

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(B) To determine the type of quadrilateral,we calculate the lengths of all four sides and the two diagonals using the distance formula: $d = \sqrt{(x_

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Rectangle

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(B) To determine the type of quadrilateral,we calculate the lengths of all four sides and the two diagonals using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Sides:$AB = \sqrt{(7 - 2)^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$CD = \sqrt{(6 - 11)^2 + (-6 - (-1))^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$DA = \sqrt{(2 - 6)^2 + (-2 - (-6))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$Diagonals:$AC = \sqrt{(11 - 2)^2 + (-1 - (-2))^2} = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82}$$BD = \sqrt{(6 - 7)^2 + (-6 - 3)^2} = \sqrt{(-1)^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}$Since opposite sides are equal ($AB = CD$ and $BC = DA$) and the diagonals are equal $(AC = BD)$,the quadrilateral is a rectangle.

What type of a quadrilateral do the points $A(2, -2)$,$B(7, 3)$,$C(11, -1)$,and $D(6, -6)$ taken in that order,form?

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