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(A) Let the vertices of the square be $A(a, b)$ and $C(b, a)$.These are opposite vertices,so the distance $AC$ is the length of the diagonal $d$ of th

Vedclass · Accepted Answer

$(a - b)^2$

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(A) Let the vertices of the square be $A(a, b)$ and $C(b, a)$.These are opposite vertices,so the distance $AC$ is the length of the diagonal $d$ of the square.The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$d = \sqrt{(b - a)^2 + (a - b)^2} = \sqrt{(a - b)^2 + (a - b)^2} = \sqrt{2(a - b)^2} = |a - b|\sqrt{2}$.The area of a square in terms of its diagonal $d$ is given by $	ext{Area} = \frac{1}{2}d^2$.$	ext{Area} = \frac{1}{2} 	imes (|a - b|\sqrt{2})^2 = \frac{1}{2} 	imes (a - b)^2 	imes 2 = (a - b)^2$.

If the coordinates of two opposite vertices of a square are $(a, b)$ and $(b, a)$,find the area of the square.

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