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A variable line \mathrm{' L '} is drawn through \mathrm{\mathrm{O}(0,0)} to meet the lines \mathrm{\mathrm{L} 1: \mathrm{y}-\mathrm{x}-10=0} and \mathrm{L 2: y-x-20=0} at the points \mathrm{A\: and \: B} respectively. A point \mathrm{ P} is taken on \mathrm{' L '} such that \mathrm{ \frac{2}{O P}=\frac{1}{O A}+\frac{1}{\mathbf{O B}}}. Locus of \mathrm{ ' \mathrm{P} '} is -
 

Option: 1

\mathrm{3 x+3 y=40}


 


Option: 2

\mathrm{3 x+3 y+40=0}
 


Option: 3

\mathrm{3 x-3 y=40}
 


Option: 4

\mathrm{3 y-3 x=40}


Answers (1)

best_answer

Let the parametric equation of drawn line is \mathrm{\frac{x}{\cos \theta}=\frac{y}{\sin \theta}=\mathrm{r}}

\mathrm{\Rightarrow \mathrm{x}=\mathrm{r} \cos \theta, \mathrm{y}=\mathrm{r} \sin \theta}

Putting it in \mathrm{' \mathrm{L} 1 '} we get \mathrm{\mathrm{r} \sin \theta=r \cos \theta+10}

\mathrm{ \Rightarrow \frac{1}{O A}=\frac{\sin \theta-\cos \theta}{10} }

Similarly putting the general point of drawn line in the equation of \mathrm{ L2 }, we get

\mathrm{ \frac{1}{O B}=\frac{\sin \theta-\cos \theta}{20} }

Let \mathrm{ P=(h 1 k)\: and\: O P=r \Rightarrow r \cos \theta=h r \sin \theta=k }, we have

\mathrm{ \frac{2}{r}=\frac{\sin \theta-\cos \theta}{10}+\frac{\sin \theta-\cos \theta}{20} }

\mathrm{ \Rightarrow 40=3 \mathrm{r} \sin \theta-3 \mathrm{r} \cos \theta }

\mathrm{ \Rightarrow 3 \mathrm{y}-3 \mathrm{x}=40 }

Hence option 4 is correct.



 

Posted by

Ritika Kankaria

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