A variable line ' L ' is drawn through $$O(0$$,$$0)$$ to meet the lines $$L1$$:y−x−$$10=0$$ and $$L2$$:y−x−$$20=0$$ at the points A and B respectively. A point P is taken on ' L′ such that $$2OP=1OA+1OB$$ . Locus of ' P′ is

Question

A variable line ' L ' is drawn through $$O(0$$,$$0)$$ to meet the  lines $$L1$$:y−x−$$10=0$$ and $$L2$$:y−x−$$20=0$$ at the  points A and B respectively. A point P is taken on ' L′ such that $$2OP=1OA+1OB$$ . Locus of ' P′ is

Infinity Learn · Accepted Answer

Let the parametric equation of drawn line isxcos⁡θ$$=ysin$$⁡θ$$=r$$⇒ $$x=rcos$$⁡θ,$$y=rsin$$⁡θ Putting it in ' $$L1$$ ', we get rsin⁡θ$$=rcos$$⁡θ$$+10$$⇒ $$1OA=$$$$sin$$⁡θ−$$cos$$⁡θ$$10Similarly$$, putting the general point of drawn line is the equation of $$L2$$, we get $$1OB=$$$$sin$$⁡θ−$$cos$$⁡θ$$20$$ Let $$P=(h$$,$$k)$$ and $$OP=r$$⇒rcos⁡θ$$=h$$,rsin⁡θ$$=k$$ , we have $$2r=$$$$sin$$⁡θ−$$cos$$⁡θ$$10+$$$$sin$$⁡θ−$$cos$$⁡θ$$20$$⇒ $$40=3rsin$$⁡θ−$$3rcos$$⁡θ⇒ $$3y$$−$$3x=40$$

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

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A variable line ' L ' is drawn through O(0,0) to meet the lines L1:y−x−10=0 and L2:y−x−20=0 at the points A and B respectively. A point P is taken on ' L′ such that 2OP=1OA+1OB . Locus of ' P′ is

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Ready to Test Your Skills?

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