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(B) Let the vertices of the parallelogram $ABCD$ be $A(-1, 0), B(3, 1), C(2, 2),$ and $D(x, y).$Since the diagonals of a parallelogram bisect each oth

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$(-2, 1)$

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(B) Let the vertices of the parallelogram $ABCD$ be $A(-1, 0), B(3, 1), C(2, 2),$ and $D(x, y).$Since the diagonals of a parallelogram bisect each other,the midpoint of $AC$ is the same as the midpoint of $BD.$Midpoint of $AC = (\frac{-1+2}{2}, \frac{0+2}{2}) = (\frac{1}{2}, 1).$Midpoint of $BD = (\frac{3+x}{2}, \frac{1+y}{2}).$Equating the midpoints: $(\frac{3+x}{2}, \frac{1+y}{2}) = (\frac{1}{2}, 1).$For the $x$-coordinate: $\frac{3+x}{2} = \frac{1}{2}$ $\Rightarrow 3+x = 1$ $\Rightarrow x = -2.$For the $y$-coordinate: $\frac{1+y}{2} = 1$ $\Rightarrow 1+y = 2$ $\Rightarrow y = 1.$Thus,the fourth vertex is $(-2, 1).$

If the three vertices of a parallelogram are $(-1, 0), (3, 1),$ and $(2, 2)$ respectively,find the coordinates of the fourth vertex.

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