If A (3,0), B (0,0), C (1,3) and D (4,3), the | vedclass

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(C) First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sq

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parallelogram

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(C) First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.$AB = \sqrt{(0-3)^2 + (0-0)^2} = \sqrt{9} = 3$.$BC = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.$CD = \sqrt{(4-1)^2 + (3-3)^2} = \sqrt{3^2 + 0^2} = 3$.$AD = \sqrt{(4-3)^2 + (3-0)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.Since opposite sides are equal ($AB = CD = 3$ and $BC = AD = \sqrt{10}$),the quadrilateral is a parallelogram.Next,we check the diagonals:$AC = \sqrt{(1-3)^2 + (3-0)^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$.$BD = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$.Since the diagonals are not equal $(AC 
eq BD)$,it is not a rectangle.Therefore,$\square ABCD$ is a parallelogram.

If $A (3,0), B (0,0), C (1,3)$ and $D (4,3),$ then $\square ABCD$ is a $\ldots \ldots \ldots \ldots$

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