Three vertices of a parallelogram taken in or | vedclass

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(B) Let the vertices of the parallelogram be $A(-1, -6)$,$B(2, -5)$,$C(7, 2)$,and $D(x, y)$.Since the diagonals of a parallelogram bisect each other,t

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$(4, 1)$

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(B) Let the vertices of the parallelogram be $A(-1, -6)$,$B(2, -5)$,$C(7, 2)$,and $D(x, y)$.Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.The midpoint of $AC$ is $(\frac{-1+7}{2}, \frac{-6+2}{2}) = (3, -2)$.The midpoint of $BD$ is $(\frac{2+x}{2}, \frac{-5+y}{2})$.Equating the midpoints:$\frac{2+x}{2} = 3$ $\Rightarrow 2+x = 6$ $\Rightarrow x = 4$$\frac{-5+y}{2} = -2$ $\Rightarrow -5+y = -4$ $\Rightarrow y = 1$Therefore,the fourth vertex is $(4, 1)$.

Three vertices of a parallelogram taken in order are $(-1, -6)$,$(2, -5)$ and $(7, 2)$. The fourth vertex is

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