If $P , Q$ are two points on the line $3 x + 4 y + 15 = 0$ such that $O P = O Q = 9$ then the area of $\Delta O P Q$ is

Given in $\triangle \mathrm{OPQ}, \mathrm{O}$ is the origin and $\mathrm{OP}=\mathrm{OQ}=9$

Therefore, $\triangle \mathrm{OPQ}$ is isosceles

Let, $\mathrm{c}$ is the midpoint of $\mathrm{PQ}, \mathrm{oc}$ is the perpendiular to the line $\mathrm{PQ}$.

oc = Length of the perpendicular from $(0,0)$ to

the line $3 x+4 y+15=0$

$\Rightarrow O C=\frac{|0+0+15|}{\sqrt{3^{2}+4^{2}}}=\frac{15}{\sqrt{25}}=\frac{15}{5}=3$

In right angled triangle оср,

$\begin{array}{l} \mathrm{OP}^{2}=\mathrm{OC}^{2}+\mathrm{PC}^{2} \\ \Rightarrow 9^{2}=3^{2}+\mathrm{PC}^{2} \\ \Rightarrow \mathrm{PC}^{2}=81-9 \\ \Rightarrow \mathrm{PC}^{2}=72 \\ \Rightarrow \mathrm{PC}=\sqrt{72}=6 \sqrt{2} \end{array}$

Since, $\mathrm{c}$ is the midpoint of $\mathrm{PQ}$

$\begin{array}{l} \therefore \mathrm{PQ}=2 \times 6 \sqrt{2}=12 \sqrt{2} \\ \begin{aligned} \therefore \text { Area of } & \triangle \mathrm{OPQ}=\frac{1}{2} \times \mathrm{OC} \times \mathrm{PQ} \\ & =\frac{1}{2} \times 3 \times 12 \sqrt{2} \\ & =18 \sqrt{2} \text { square units. } \end{aligned} \end{array}$