Question

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.

Answer 1

Given three vertices of a parallelogram $ABCD$ are $A(3,-1,2),B(1,2,-4)$ and $C(-1,1,2)$ .

Let the coordinates of the fourth vertex be $D(x,y,z)$.

We know that the diagonals of a parallelogram bisect each other.

Therefore in parallelogram $ABCD$, $AC$ and $BD$ bisect each other .

$\therefore$ Mid-point of $AC=$ Mid-point of $BD$

$\Rightarrow (\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2})=(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2})$
$\Rightarrow (1,0,2)=(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2})$
$\Rightarrow \frac{x+1}{2}=1,\frac{y+2}{2}=0$ and $\frac{z-4}{2}=2$
$\Rightarrow x=1,y=-2$ and $z=8$

Thus, the coordinates of the fourth vertex are $(1,-2,8)$