Question: The equation of the line which bisects the obtuse angle between the lines $ x-2y+4=0 $ and $ 4x-3y+2=0 $ , is [IIT 1979]
Options:
A) $ (4-\sqrt{5})x-(3-2\sqrt{5})y+(2-4\sqrt{5})=0 $
B) $ (4+\sqrt{5})x-(3+2\sqrt{5})y+(2+4\sqrt{5})=0 $
C) $ (4+\sqrt{5})x+(3+2\sqrt{5})y+(2+4\sqrt{5})=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The equations of the bisectors of the angles between the lines are $ \frac{x-2y+4}{\sqrt{1+4}}=\pm \frac{4x-3y+2}{\sqrt{16+9}} $ Taking positive sign, then $ (4-\sqrt{5})x-(3-2\sqrt{5})y-(4\sqrt{5}-2)=0 $ …..(i) and negative sign gives $ (4+\sqrt{5})x-(2\sqrt{5}+3)y+(4\sqrt{5}+2)=0 $ …..(ii) Let $ \theta $ be the angle between the line (i) and one of the given line, then $ \tan \theta =| \frac{\frac{1}{2}-\frac{4-\sqrt{5}}{3-2\sqrt{5}}}{1+\frac{1}{2}.\frac{4-\sqrt{5}}{3-2\sqrt{5}}} |=\sqrt{5}+2>1 $ Hence the line (i) bisects the obtuse angle between the given lines.
BETA
