Missing coordinates of a parallelogram

Question

We are given points $A(6,2,4)$, $B(7,5,9)$, $C(x,6,13)$ and $D(8,3,z)$. I need to find $x$ and $z4 so $ABCD$ is a parallelogram. I know there has been similar questions but I cant seem to get it to work. Which properties should I use and how?

Answer 1

Use $$\overrightarrow{AB} = \overrightarrow{DC} \;\;\;\Longrightarrow \;\;\; C-D = B-A$$ so $$ (x,6,13)-(8,3,z) = (7,5,9)-(6,2,4)$$ thus

$$x-8= 7-6 \;\;\;\Longrightarrow \;\;\; x= 9$$ and $$ 13-z = 9-4 \;\;\;\Longrightarrow \;\;\; z=8$$

Answer 2

The properties should be quoted is "the diagonals of a //gm bisect each other". Therefore, x and z can be found by equating the midpoints of (A and C) and (B and D).