Three consecutive vertices of a parallelogram
Question:

Three consecutive vertices of a parallelogram ABCD are A (6, – 2, 4), B (2, 4, – 8), C (–2, 2, 4). Find the coordinates of the fourth vertex.

Solution:

Given three consecutive vertices of a parallelogram $A B C D$ are $A(6,-2,4), B(2,$, $4,-8), C(-2,2,4)$.

Let the forth vertex be $\mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z})$.

Midpoint of diagonal $A C=P\left(\frac{6-2}{2}, \frac{-2+2}{2}, \frac{4+4}{2}\right)=P(2,0,4)$

Midpoint of diagonal $\mathrm{BD}=\mathrm{P}\left(\frac{\mathrm{x}+2}{2}, \frac{\mathrm{y}+4}{2}, \frac{\mathrm{z}-\mathrm{8}}{2}\right)=\mathrm{P}(2,0,4)$

$\Rightarrow \frac{x+2}{2}=2 \Rightarrow x=2$

$\Rightarrow \frac{y+4}{2}=0 \Rightarrow y=-4$

$\Rightarrow \frac{\mathrm{Z}-8}{2}=4 \Rightarrow \mathrm{Z}=16$

$\therefore \mathrm{D}(2,-4,16)$ is the fourth vertex.

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