SATHEE: Straight Line Question 35

Straight Line Question 35

Question: The bisector of the acute angle formed between the lines $ 4x-3y+7=0 $ and $ 3x-4y+14=0 $ has the equation [Pb. CET 2004]

Options:

A) $ x+y+3=0 $

B) $ x-y-3=0 $

C) $ x-y+3=0 $

D) $ 3x+y-7=0 $

Show Answer

Answer:

Correct Answer: C

Solution:

The equation of bisector of acute angle formed between the lines $ 4x-3y+7=0 $ and $ 3x-4y+14=0 $ is $ \frac{4x-3y+7}{\sqrt{16+9}}=-\frac{3x-4y+14}{\sqrt{16+9}} $
Þ $ 7x-7y+21=0 $
Þ $ x-y+3=0 $ .

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Straight Line Question 35

Question: The bisector of the acute angle formed between the lines $ 4x-3y+7=0 $ and $ 3x-4y+14=0 $ has the equation [Pb. CET 2004]

Options:

Answer:

Solution:

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