Which of the following statements are true?

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(A) For option $A$: The midpoint of diagonal $AC$ is $(\frac{-2+4}{2}, \frac{-1+3}{2}) = (1, 1)$. The midpoint of diagonal $BD$ is $(\frac{1+1}{2}, \f

Vedclass · Accepted Answer

$A(-2, -1), B(1, 0), C(4, 3)$ and $D(1, 2)$ are the vertices of a parallelogram.

Vedclass · Answer

(A) For option $A$: The midpoint of diagonal $AC$ is $(\frac{-2+4}{2}, \frac{-1+3}{2}) = (1, 1)$. The midpoint of diagonal $BD$ is $(\frac{1+1}{2}, \frac{0+2}{2}) = (1, 1)$. Since the midpoints of the diagonals coincide,the diagonals bisect each other,so $ABCD$ is a parallelogram.For option $B$: Calculating side lengths,$AB = \sqrt{(-2 - (-4))^2 + (-4 - (-1))^2} = \sqrt{2^2 + (-3)^2} = \sqrt{13}$. $BC = \sqrt{(4 - (-2))^2 + (0 - (-4))^2} = \sqrt{6^2 + 4^2} = \sqrt{52}$. Since $AB 
eq BC$,it is not a square.For option $C$: Calculating side lengths,$AB = \sqrt{(2-0)^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8}$. $BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$. $CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8}$. $DA = \sqrt{(0 - (-2))^2 + (-1-1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8}$. All sides are equal. The diagonal $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = 4$ and $BD = \sqrt{(-2-2)^2 + (1-1)^2} = 4$. Since diagonals are equal and sides are equal,it is a square,not a rhombus.Thus,only statement $A$ is true.

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