<s>[INST] Find the area of $\triangle POQ$ . 
If $P$ and $Q$ are two points on the line $3x+4y=-15$ such that $OP=OQ=9$, then the area of $\triangle POQ$ will be ?

$\color{green}{a.)18\sqrt2}\\
b.)3\sqrt2\\
c.)6\sqrt2\\
d.)15\sqrt2$
The information about $O$ is not given ,
I assumed it to be origin $(0,0)$ and by solving $x^2+y^2=9$ and $3x+4y=-15$
and  got only $\left(-\dfrac{9}{5},-\dfrac{12}{5}\right)$.
I am not getting ideas
I look for a short method.
I have studied maths upto $12th$ grade.
 [/INST] $\Delta POQ$ is an isosceles triangle with $OP=OQ=9$. Now, draw the perpendicular OM from the point O to the line: $3x+4y=-15$. Then we get $$OM=\frac{|3(0)+4(0)+15|}{\sqrt{3^2+4^2}}=\frac{15}{5}=3$$ Now, in right $\Delta OMP$, we have $$PM=\sqrt{(OP)^2-(OM)^2}=\sqrt{(9)^2-(3)^2}=6\sqrt{2}$$ Also, in $\Delta POQ$ we have $PQ=2(PM)=12\sqrt{2}$ Hence, the area of isosceles $\Delta POQ$ $$=\frac{1}{2}(OM)(PQ)=\frac{1}{2}(3)(12\sqrt{2})=18\sqrt{2} \space \text{sq. units}$$ 
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