jmc

The area of $\square OPQR$ is $2^2=4$. The area of $△PQT$ is $\frac{1}{2}bh=\frac{1}{2}\cdot PT\cdot PQ$. If we set the areas equal, the area of $△PQT$ 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 $\boxed{(-2,0)}$.