If three vertices of a parallelogram ABCD are | vedclass

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(A) Let the coordinates of $D$ be $(x, y)$.In a parallelogram,the diagonals bisect each other,meaning the midpoint of diagonal $AC$ is the same as the

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$(2, -3)$

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(A) Let the coordinates of $D$ be $(x, y)$.In a parallelogram,the diagonals bisect each other,meaning the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The midpoint of $AC = \left(\frac{1+3}{2}, \frac{2+(-4)}{2}\right) = \left(\frac{4}{2}, \frac{-2}{2}\right) = (2, -1)$.The midpoint of $BD = \left(\frac{2+x}{2}, \frac{1+y}{2}\right)$.Equating the midpoints: $(2, -1) = \left(\frac{2+x}{2}, \frac{1+y}{2}\right)$.For the $x$-coordinate: $\frac{2+x}{2} = 2 \implies 2+x = 4 \implies x = 2$.For the $y$-coordinate: $\frac{1+y}{2} = -1 \implies 1+y = -2 \implies y = -3$.Therefore,the coordinates of $D$ are $(2, -3)$.

If three vertices of a parallelogram $ABCD$ are $A(1, 2)$,$B(2, 1)$,and $C(3, -4)$,then the coordinates of $D$ are.............

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