Straight Line Question 233
Question: The line $ x+y=a $ meets the axes of x and y at A and B respectively. A $ \Delta AMN $ is inscribed in the $ \Delta OAB,O $ being the origin, with right angle at N. M and N lie respectively on OB and AB. If the area of the $ \Delta AMN $ is $ \frac{3}{8} $ of the area of the $ \Delta OAB, $ then $ \frac{AN}{BN} $ is equal to
Options:
A) $ \frac{1}{3} $
B) $ \frac{1}{3},3 $
C) $ \frac{2}{3},3 $
D) 3
Correct Answer: D $ \Rightarrow x-y=\frac{1-\lambda }{1+\lambda }a $
So the coordinates of M are $ ( 0,\frac{\lambda -1}{\lambda +1}a ). $
Therefore, area of the $ \Delta AMN $ is
$ =\frac{1}{2}| [ a( \frac{-a}{\lambda +1} )+\frac{1-\lambda }{{{(1+\lambda )}^{2}}}a^{2} ] | $ $ =\frac{\lambda a^{2}}{{{(1+\lambda )}^{2}}} $
Also, area of $ \Delta OAB=\frac{a^{2}}{2} $
So, that according to the given condition
$ \frac{\lambda a^{2}}{{{(1+\lambda )}^{2}}}=\frac{3}{8}.\frac{1}{2}a^{2} $ $ \Rightarrow 3{{\lambda }^{2}}-10\lambda +3=0 $ $ \Rightarrow \lambda =3or\lambda =\frac{1}{3} $
For $ \lambda =\frac{1}{3},M $ lies outside the segment OB and
Hence the required value of $ \lambda $ is 3.Show Answer
Answer:
Solution:
BETA
