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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.

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  • $\begingroup$ @AnuragA That seems to be a rather safe assumption in most basic geometry books I know. $\endgroup$ – Timbuc May 26 '15 at 19:44
  • $\begingroup$ If O is not given then area will become variable since we know that you can construct isosceles triangle with various angles on the line.Maybe they forgot to mention o as origin . $\endgroup$ – Shubham Ugare May 26 '15 at 19:46
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    $\begingroup$ The equation of circle should be taken as x^2+y^2=81..not 9.. $\endgroup$ – Shubham Ugare May 26 '15 at 19:48
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$\Delta POQ$ 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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    $\begingroup$ $\quad \quad$ thnks! $\endgroup$ – R K May 26 '15 at 20:00
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Let $l$ be the line with equation $3x+4y+15=0$. If $P,Q\in l$ and $OP=OQ=9$, $$ [POQ] = \frac{1}{2} PQ\cdot d(O,l) = d(O,l)\cdot\sqrt{9^2-d(O,l)^2}\tag{1}$$ and assuming $O$ is the origin, through a well-known formula, $$ d(O,l) = \frac{15}{\sqrt{3^2+4^2}} = 3, \tag{2}$$ from which $[POQ] = \color{red}{18\sqrt{2}}$ follows.

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    $\begingroup$ $\quad \quad$ thnks! $\endgroup$ – R K May 26 '15 at 20:00

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