Straight Line Question 231
Question: The bisector of the acute angle formed between the lines $ 4x-3y+7=0 $ and $ 3x-4y+14=0 $ has the equation:
Options:
A) $ x+y+3=0 $
B) $ x-y-3=0 $
C) $ x-y+3=0 $
D) $ 3x+y-7=0 $
Correct Answer: C $ \Rightarrow 4h-3k+7=\pm (3h-4k+14) $ $ \Rightarrow h+k-7=0 $ and $ 7h-7k+21=0 $ Hence locus of $ (h,k) $ is $ x+y-7=0 $ and $ x-y+3=0 $
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Answer:
Solution:
BETA
