Question

Given that the points A=(-1,4,0), B=(2,1,1), C=(2,-2,1) are three vertices of the parallelogram ABCD, findthe fourth vertex D.

Answer 1

GIVEN

A parallelogram ABCD with given vertices

$A=\left(-1,4,0\right),B=\left(2,1,1\right),C=\left(2,-2,1\right)$

SOLUTION

Let the fourth vertex D have coordinates (x, y, z).

The diagonals of a parallelogram bisect each other. Therefore, the midpoint of the two diagonals must coincide.

For a parallelogram ABCD, the diagonals are AC and BD.

Recall the midpoint formula:

$(x_m,y_m,z_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$

Therefore, the midpoint for AC will be:

$AC\Rightarrow(\frac{-1+2}{2},\frac{4+(-2)}{2},\frac{0+1}{2})=(0.5,1,0.5)$

The midpoint of BD will be:

$BD\Rightarrow(\frac{2+x}{2},\frac{1+y}{2},\frac{1+z}{2})$

Equate the midpoints:

$(\frac{2+x}{2},\frac{1+y}{2},\frac{1+z}{2})=(0.5,1,0.5)$

Therefore:

$\begin{gathered} \frac{2+x}{2}=0.5,x=-1 \\ \frac{1+y}{2}=1,y=1 \\ \frac{1+z}{2}=0.5,z=0 \end{gathered}$

Therefore, the coordinates of the fourth vertex will be:

$(x,y,z)=(-1,1,0)$