A) \[x+y+3=0\]
B) \[x-y-3=0\]
C) \[x-y+3=0\]
D) \[3x+y-7=0\]
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 \[x-y+3=0\].You need to login to perform this action.
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