The quadrilateral formed by the vertices A(-1 | vedclass

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(B) Let the vertices be $A(-1, 1)$,$B(0, -3)$,$C(5, 2)$,and $D(4, 6)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 -

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Parallelogram

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(B) Let the vertices be $A(-1, 1)$,$B(0, -3)$,$C(5, 2)$,and $D(4, 6)$.Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(0 - (-1))^2 + (-3 - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}$$BC = \sqrt{(5 - 0)^2 + (2 - (-3))^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}$$CD = \sqrt{(4 - 5)^2 + (6 - 2)^2} = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}$$DA = \sqrt{(-1 - 4)^2 + (1 - 6)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}$Since opposite sides are equal ($AB = CD = \sqrt{17}$ and $BC = DA = \sqrt{50}$),the quadrilateral is a parallelogram.Check the diagonals:$AC = \sqrt{(5 - (-1))^2 + (2 - 1)^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$$BD = \sqrt{(4 - 0)^2 + (6 - (-3))^2} = \sqrt{4^2 + 9^2} = \sqrt{16 + 81} = \sqrt{97}$Since $AC 
eq BD$,it is not a rectangle or a square.Thus,the quadrilateral is a parallelogram.

The quadrilateral formed by the vertices $A(-1, 1)$,$B(0, -3)$,$C(5, 2)$,and $D(4, 6)$ is:

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