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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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