A variable line L passing through the origin | vedclass

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(C) Let the equation of the line passing through the origin be $y=mx$,which cuts the parallel lines $x-y+10=0$ and $x-y+20=0$ at points $A$ and $B$ re

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$3x-3y+40=0$

Vedclass · Answer

(C) Let the equation of the line passing through the origin be $y=mx$,which cuts the parallel lines $x-y+10=0$ and $x-y+20=0$ at points $A$ and $B$ respectively. For point $A$: $x-mx+10=0$ $\Rightarrow x(1-m)=-10$ $\Rightarrow x=\frac{10}{m-1}, y=\frac{10m}{m-1}$. Thus,$OA = \sqrt{x^2+y^2} = \sqrt{\frac{100(1+m^2)}{(m-1)^2}} = \frac{10\sqrt{1+m^2}}{|m-1|}$. Similarly,for point $B$: $x-mx+20=0 \Rightarrow x=\frac{20}{m-1}, y=\frac{20m}{m-1}$. Thus,$OB = \frac{20\sqrt{1+m^2}}{|m-1|}$. Let $P(h, k)$ be a point on $y=mx$,so $m=\frac{k}{h}$ and $OP = \sqrt{h^2+k^2}$. Since $OA, OP, OB$ are in harmonic progression,$\frac{2}{OP} = \frac{1}{OA} + \frac{1}{OB}$. Substituting the values: $\frac{2}{\sqrt{h^2+k^2}} = \frac{|m-1|}{10\sqrt{1+m^2}} + \frac{|m-1|}{20\sqrt{1+m^2}} = \frac{|m-1|}{\sqrt{1+m^2}} \left(\frac{1}{10} + \frac{1}{20}\right) = \frac{|m-1|}{\sqrt{1+m^2}} \cdot \frac{3}{20}$. Since $m = \frac{k}{h}$,$\sqrt{1+m^2} = \sqrt{1+\frac{k^2}{h^2}} = \frac{\sqrt{h^2+k^2}}{|h|}$. So,$\frac{2}{\sqrt{h^2+k^2}} = \frac{|\frac{k}{h}-1|}{\frac{\sqrt{h^2+k^2}}{|h|}} \cdot \frac{3}{20} = \frac{|k-h|}{\sqrt{h^2+k^2}} \cdot \frac{3}{20}$. $40 = 3|k-h| \Rightarrow 3(k-h) = \pm 40$. Given the geometry,the locus is $3x-3y+40=0$.

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

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