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).