$\text{ A variable line ' }L\text{ ' is drawn through }O(0,0)\text{ to meet the lines }{L}_1:y-x-10=0\text{ and }{L}_2:y-x-20=0\text{ at the }$

$\text{ points }A\text{ and }B\text{ respectively. A point }P\text{ is taken on ' }{L}^{\mathrm{\prime }}\text{ such that }\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}\text{ . Locus of ' }{P}^{\mathrm{\prime }}\text{ is }$

Let the parametric equation of drawn line is

$\begin{array}{r}\frac{x}{\cos \theta }=\frac{y}{\sin \theta }=r\\ \Rightarrow x=r\cos \theta ,y=r\sin \theta \end{array}$

$\begin{array}{r}\text{ Putting it in ' }{L}_1\text{ ', we get }\\ r\sin \theta =r\cos \theta +10\\ \Rightarrow \frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10}\end{array}$

Similarly, putting the general point of drawn line is the equation of L₂, we get 

$\frac{1}{OB}=\frac{\sin \theta -\cos \theta }{20}$

$\begin{array}{l}\text{ Let }P=(h,k)\text{ and }OP=r\Rightarrow r\cos \theta =h,r\sin \theta =k\text{ , we have }\\ \frac{2}{r}=\frac{\sin \theta -\cos \theta }{10}+\frac{\sin \theta -\cos \theta }{20}\\ \Rightarrow 40=3r\sin \theta -3r\cos \theta \\ \Rightarrow 3y-3x=40\end{array}$