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Question

The equation of the bisector of the acute angle between 4x+3y−6=0 and 5x+12y+9=0 is

A
9x7y41=0
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B
7x+9y3=0
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C
9x+7y3=0
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D
9x7y+41=0
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Solution

The correct option is B 7x+9y3=0
Given lines are 4x+3y6=0 and 5x+12y+9=0
Making the constant terms positive, we get
4x3y+6=0 and 5x+12y+9=0
Now,
a1a2+b1b2=2036=56<0
So, the equation of acute angle bisector is
4x3y+6(4)2+(3)2=5x+12y+9(5)2+(12)24x3y+65=5x+12y+913
13(4x3y+6)=5(5x+12y+9)52x39y+78=25x+60y+4577x+99y33=07x+9y3=0

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