Straight Line Question 30
Question: The equation of the bisector of the acute angle between the lines $ 3x-4y+7=0 $ and $ 12x+5y-2=0 $ is [IIT 1975, 1983; RPET 2003; UPSEAT 2004]
Options:
A) $ 21x+77y-101=0 $
B) $ 11x-3y+9=0 $
C) $ 31x+77y+101=0 $
D) $ 11x-3y-9=0 $
Correct Answer: B $ \Rightarrow \alpha <45^{o} $ Hence $ 11x-3y+9=0 $ is the bisector of the acute angle between the given lines.
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Answer:
Solution:
Þ $ 11x-3y+9=0 $ ……(i) and $ 21x+77y-101=0 $ ……(ii) Let the angle between the line $ 3x-4y+7=0 $ and (i) is $ \alpha , $ then $ \tan \alpha =| \frac{m_1-m_2}{1+m_1m_2} |=| \frac{\frac{3}{4}-\frac{11}{3}}{1+\frac{3}{4}\times \frac{11}{3}} |=\frac{35}{45}<1 $
BETA
