Three vertices of a parallelogram ABCD are A( | vedclass

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(A) Let the coordinates of the fourth vertex $D$ be $(x, y, z)$.In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same mi

Vedclass · Accepted Answer

$(1, -2, 8)$

Vedclass · Answer

(A) Let the coordinates of the fourth vertex $D$ be $(x, y, z)$.In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.Midpoint of $AC = \left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right) = \left(\frac{2}{2}, \frac{0}{2}, \frac{4}{2}\right) = (1, 0, 2)$.Midpoint of $BD = \left(\frac{x + 1}{2}, \frac{y + 2}{2}, \frac{z - 4}{2}\right)$.Equating the midpoints:$\frac{x + 1}{2} = 1$ $\Rightarrow x + 1 = 2$ $\Rightarrow x = 1$.$\frac{y + 2}{2} = 0$ $\Rightarrow y + 2 = 0$ $\Rightarrow y = -2$.$\frac{z - 4}{2} = 2$ $\Rightarrow z - 4 = 4$ $\Rightarrow z = 8$.Thus,the coordinates of the fourth vertex $D$ are $(1, -2, 8)$.

Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$,$B(1, 2, -4)$,and $C(-1, 1, 2)$. Find the coordinates of the fourth vertex $D$.

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