The opposite angular points of a square are ( | vedclass

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(C) Let the given opposite vertices be $A(3, 4)$ and $C(1, -1)$. The midpoint of $AC$ is $M = \left( \frac{3+1}{2}, \frac{4-1}{2} \right) = \left( 2,

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$D \left( \frac{9}{2}, \frac{1}{2} \right), B \left( -\frac{1}{2}, \frac{5}{2} \right)$

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(C) Let the given opposite vertices be $A(3, 4)$ and $C(1, -1)$. The midpoint of $AC$ is $M = \left( \frac{3+1}{2}, \frac{4-1}{2} \right) = \left( 2, \frac{3}{2} \right)$.Let the other two vertices be $B$ and $D$. In a square,the diagonals are equal and bisect each other at right angles.The vector $\vec{AC} = (1-3, -1-4) = (-2, -5)$.The vector $\vec{BD}$ must be perpendicular to $\vec{AC}$ and have the same length. $A$ vector perpendicular to $(-2, -5)$ is $(5, -2)$ or $(-5, 2)$.Since the diagonals bisect each other,the vertices $B$ and $D$ are given by $M \pm \frac{1}{2} \vec{BD}'$,where $\vec{BD}'$ is a vector of length equal to $AC$ perpendicular to $AC$.The length $AC = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4+25} = \sqrt{29}$.The vector $\vec{BD}'$ has length $\sqrt{29}$ and is perpendicular to $\vec{AC}$. Let $\vec{BD}' = (5, -2)$.Then $B, D = M \pm \frac{1}{2} (5, -2) = \left( 2 \pm \frac{5}{2}, \frac{3}{2} \mp 1 \right)$.For the plus sign: $\left( 2 + \frac{5}{2}, \frac{3}{2} - 1 \right) = \left( \frac{9}{2}, \frac{1}{2} \right)$.For the minus sign: $\left( 2 - \frac{5}{2}, \frac{3}{2} + 1 \right) = \left( -\frac{1}{2}, \frac{5}{2} \right)$.Thus,the coordinates are $\left( \frac{9}{2}, \frac{1}{2} \right)$ and $\left( -\frac{1}{2}, \frac{5}{2} \right)$,which matches option $C$.

The opposite angular points of a square are $(3, 4)$ and $(1, -1)$. Then the coordinates of the other two points are

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