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(C) Let the vertices of the rectangle be $A(2, -2)$,$B(8, 4)$,$C(5, 7)$,and $D(x, y)$.Since the diagonals of a rectangle bisect each other,the midpoin

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$(-1, 1)$

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(C) Let the vertices of the rectangle be $A(2, -2)$,$B(8, 4)$,$C(5, 7)$,and $D(x, y)$.Since the diagonals of a rectangle bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.Midpoint of $AC = (\frac{2 + 5}{2}, \frac{-2 + 7}{2}) = (3.5, 2.5)$.Midpoint of $BD = (\frac{8 + x}{2}, \frac{4 + y}{2})$.Equating the midpoints:$\frac{8 + x}{2} = 3.5$ $\Rightarrow 8 + x = 7$ $\Rightarrow x = -1$.$\frac{4 + y}{2} = 2.5$ $\Rightarrow 4 + y = 5$ $\Rightarrow y = 1$.Thus,the fourth vertex is $(-1, 1)$.

If the three vertices of a rectangle taken in order are the points $(2, -2)$,$(8, 4)$,and $(5, 7)$,then the coordinates of the fourth vertex are:

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