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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 $4^{th}$ vertex $D$.

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  • Section formula: The coordinates of the point $C$ that divides the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $1:1$ internally is given by $C\big(\large\frac{x_2+x_1}{2},\frac{y_2+y_1}{2},\frac{z_2+z_1}{2}\big)$
Given three vertices of a parallelogram are
Let the $4^{th}$ vertex $D$ be $D(x,y,z)$
In the parallelogram $ABCD$, $AC$ and $BD$ are diagonals.
We know that the diagonals of a parallelogram bisect each other.
Let the point of bisection be $E$
Using section formula $E$ divides both $BD$ and $AC$ in the ratio $1:1$
Using $AC$ the coordinates of $E$ is given by
$E\big(\large\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\big)$ $=E(1,0,2)$
Similarly using $BD$ the coordinates of $E$ is given by
Equating the coordinates of $E$,
$\therefore$ The coordinates of $D(x,y,z)=D(1,-2,8)$
answered May 11, 2014 by rvidyagovindarajan_1

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