Figure $OPQR$ is a square.  Point $O$ is the origin, and point $Q$ has coordinates (2,2).  What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$?


NOT TO SCALE
$\text{(A)}\ (-6,0) \qquad \text{(B)}\ (-4,0) \qquad \text{(C)}\ (-2,0) \qquad \text{(D)}\ (2,0) \qquad \text{(E)}\ (4,0)$
The area of $\square OPQR$ is $2^2 = 4$.
The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$.
If we set the areas equal, the area of $\triangle PQT$ is $4$.  Also, note that $PQ=2$.  Plugging those in, we get:
$4 = \frac{1}{2} \cdot PT \cdot 2$
$PT = 4$
If $PT = 4$, and $PO = 2$, then $OT = 2$, and $T$ must be $2$ units to the left of the origin.  This would be $(-2,0)$, giving answer $\boxed{C}$.