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

The three vertices of a parallelogram ABCD are given as 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 a parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
⇒ ($\frac{3-1}{2}$, $\frac{-1+1}{2}$, $\frac{2+2}{2}$) = ($\frac{x+1}{2}$, $\frac{y+2}{2}$, $\frac{z-4}{2}$)
⇒ (1,0,2) = ($\frac{x+1}{2}$, $\frac{y+2}{2}$, $\frac{z-4}{2}$)
⇒ $\frac{x+1}{2}$=1, $\frac{y+2}{2}$ =0, and $\frac{z-4}{2}$ = 2
⇒x=1, y = -2, and z = 8
Thus, the coordinates of the fourth vertex are (1, -2, 8).