A variable line ‘L’ passing through the origin cuts twoparallel line $\mathrm{x}-\mathrm{y}+10=0$ and $\mathrm{x}-\mathrm{y}+20=0$ at two points A and B. If P is a point on the line ‘L’ such that OA, OP, OB are in harmonic progression, then the locus of P is

let the equation of line passing through origin is y=mx   -   (1)

Given parallel lines x-y+10=0   -   (2)

let (1) cuts the line (2) be at A

$$\begin{gathered} \Rightarrow \mathrm{x}-\mathrm{mx}+10=0 \\ \Rightarrow \mathrm{x}(1-\mathrm{m})=-10 \\ \Rightarrow \mathrm{x}=\frac{-10}{1-\mathrm{m}}=\frac{10}{\mathrm{m}-1} \\ \therefore \mathrm{A}=\left(\frac{10}{\mathrm{m}-1}, \frac{10\mathrm{m}}{\mathrm{m}-1}\right) \\ \mathrm{Now} \mathrm{OA}=\sqrt{{\left(\frac{10}{\mathrm{m}-1}\right)}^2+{\left(\frac{10\mathrm{m}}{\mathrm{m}-1}\right)}^2}=\sqrt{\frac{100}{{(\mathrm{m}-1)}^2}(1+{\mathrm{m}}^2)}=\frac{10\sqrt{1+{\mathrm{m}}^2}}{\mathrm{m}-1} \end{gathered}$$

and (1) cuts the line  x-y+20=0 at B then

$$\begin{gathered} \Rightarrow \mathrm{x}-\mathrm{mx}+20=0 \\ \Rightarrow \mathrm{x}=\frac{20}{\mathrm{m}-1} \mathrm{and} \mathrm{y}=\frac{20\mathrm{m}}{\mathrm{m}-1} \\ \therefore \mathrm{B}=\left(\frac{20}{\mathrm{m}-1}, \frac{20\mathrm{m}}{\mathrm{m}-1}\right) \\ \mathrm{Now} \mathrm{OB}=\sqrt{\frac{400}{{(\mathrm{m}-1)}^2}+\frac{400{\mathrm{m}}^2}{{(\mathrm{m}-1)}^2}}=\frac{20\sqrt{1+{\mathrm{m}}^2}}{\mathrm{m}-1} \end{gathered}$$

Let P=(h, k) be locus point which lies on the line 

$\mathrm{y}=\mathrm{mx} \Rightarrow \mathrm{k}=\mathrm{mh} \Rightarrow \mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}}$

and $\mathrm{OP}=\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}$

Given that OA, OP, OB are in H.P

$$\begin{gathered} \Rightarrow \frac{2}{\mathrm{OP}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}} \\ \Rightarrow \frac{2}{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}=\frac{\mathrm{m}-1}{10\sqrt{1+{\mathrm{m}}^2}}+\frac{\mathrm{m}-1}{20\sqrt{1+{\mathrm{m}}^2}} \\ \Rightarrow \frac{2}{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}=\frac{\mathrm{m}-1}{10\sqrt{1+{\mathrm{m}}^2}}\left(1+\frac{1}{2}\right) \\ \mathrm{sub} \mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}} \\ \Rightarrow \frac{2}{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}=\frac{{\displaystyle \frac{\mathrm{k}}{\mathrm{h}}}-1}{10\sqrt{1+{\displaystyle \frac{{\mathrm{k}}^2}{{\mathrm{h}}^2}}}}\left(\frac{3}{2}\right) \\ \Rightarrow \frac{2}{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}=\frac{3\left({\displaystyle \frac{\mathrm{k}-\mathrm{h}}{\mathrm{h}}}\right)}{20{\displaystyle \frac{\sqrt{{\mathrm{h}}^2+{\mathrm{k}}^2}}{\mathrm{h}}}} \Rightarrow 2=\frac{3(\mathrm{k}-\mathrm{h})}{20} \\ \Rightarrow 3(\mathrm{k}-\mathrm{h})=40 \\ \Rightarrow 3\mathrm{h}-3\mathrm{k}+40=0 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{the} \mathrm{point} (\mathrm{h}, \mathrm{k}) \mathrm{is} 3\mathrm{x}-3\mathrm{y}+40=0 \end{gathered}$$