A parallelogram has vertices A(4,4,-1),B(5,6, | vedclass

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(C) Given,$ABCD$ is a parallelogram with vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$.We know that the diagonals of a parallelogram $A

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$(5,3,1)$

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(C) Given,$ABCD$ is a parallelogram with vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$.We know that the diagonals of a parallelogram $ABCD$ bisect each other.Therefore,the mid-point of $AC$ = mid-point of $BD$.$\left(\frac{4+6}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$$\left(\frac{10}{2}, \frac{9}{2}, 0\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$On comparing both sides,we get:$\frac{x+5}{2} = \frac{10}{2}$ $\Rightarrow x+5 = 10$ $\Rightarrow x = 5$$\frac{y+6}{2} = \frac{9}{2}$ $\Rightarrow y+6 = 9$ $\Rightarrow y = 3$$\frac{z-1}{2} = 0$ $\Rightarrow z-1 = 0$ $\Rightarrow z = 1$Thus,the vertex $D(x, y, z)$ is $(5, 3, 1)$.

$A$ parallelogram has vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$. Then the vertex $D$ is

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