If P, Q are two points on the line 3x+4y+15=0 such that OP =OQ=9, then the area of $∆OPQ$  is

$$\begin{gathered} OM=\perp dis\tan ce from (0 ,0) to 3x+4y+15=0 \\ =\frac{\left|15\right|}{\sqrt{3^2+4^2}}=\frac{15}{5}=3 \end{gathered}$$

$$\begin{gathered} From ∆ OPM , P{M}^2=O{P}^2-O{M}^2=81-9=72\Rightarrow PM=\sqrt{72}=6\sqrt{2} \\ \Rightarrow PQ=2PM=12\sqrt{2} \\ Area of ∆OPQ =\frac{1}{2}\times PQ\times OM=\frac{1}{2}\times 12\sqrt{2}\times 3=18\sqrt{2} \end{gathered}$$