Straight Line Question 258
Question: 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}{OP}=\frac{1}{OA}+\frac{1}{OB}. $ Locus of ?P? is
Options:
A) $ 3x+3y=40 $
B) $ 3x+3y+40=0 $
C) $ 3x-3y=40 $
D) $ 3y-3x=40 $
Correct Answer: D $ \Rightarrow \frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10} $
Similarly, putting the
general point of drawn
line is the equation of $ L_2, $
we get
$ \frac{1}{OB}=\frac{\sin \theta -\cos \theta }{20} $
Let $ P=(h,k) $ and $ OP=r $ $ \Rightarrow r\cos \theta =h,r\sin \theta =k, $ 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. $Show Answer
Answer:
Solution:
BETA
