jmc

The $y$-intercept of the line $y=-\frac{3}{4}x+9$ is $y=9$, so $Q$ has coordinates $(0,9)$.

To determine the $x$-intercept, we set $y=0$, and so obtain $0=-\frac{3}{4}x+9$ or $\frac{3}{4}x=9$ or $x=12$. Thus, $P$ has coordinates $(12,0)$.

Therefore, the area of $△POQ$ is $\frac{1}{2}(12)(9)=54$, since $△POQ$ is right-angled at $O$.

Since we would like the area of $△TOP$ to be one third that of $△POQ$, then the area of $△TOP$ should be 18.

If $T$ has coordinates $(r,s)$, then $△TOP$ has base $OP$ of length 12 and height $s$, so $\frac{1}{2}(12)(s)=18$ or $6s=18$ or $s=3$.

Since $T$ lies on the line, then $s=-\frac{3}{4}r+9$ or $3=-\frac{3}{4}r+9$ or $\frac{3}{4}r=6$ or $r=8$.

Thus, $r+s=8+3=\boxed{11}$.