$ABCD$ is a parallelogram with vertices $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$,and $C(x_{3}, y_{3})$. Find the coordinates of the fourth vertex $D$ in terms of $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}$,and $y_{3}$.

(N/A) Let the coordinates of vertex $D$ be $(x, y)$.
We know that the diagonals of a parallelogram bisect each other at their midpoint.
Therefore,the midpoint of diagonal $AC$ is equal to the midpoint of diagonal $BD$.
The midpoint of $AC$ is given by $\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)$.
The midpoint of $BD$ is given by $\left(\frac{x_{2}+x}{2}, \frac{y_{2}+y}{2}\right)$.
Equating the coordinates,we get:
$\frac{x_{1}+x_{3}}{2} = \frac{x_{2}+x}{2} \implies x_{1}+x_{3} = x_{2}+x \implies x = x_{1}+x_{3}-x_{2}$.
$\frac{y_{1}+y_{3}}{2} = \frac{y_{2}+y}{2} \implies y_{1}+y_{3} = y_{2}+y \implies y = y_{1}+y_{3}-y_{2}$.
Thus,the coordinates of vertex $D$ are $(x_{1}+x_{3}-x_{2}, y_{1}+y_{3}-y_{2})$.