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Question

Equation of the bisector of the acute angle between lines 3x+4y+5=0 and 12x5y7=0 is:

A
21x+77y+100=0
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B
99x27y+30=0
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C
99x+27y+30=0
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D
21x77y100=0
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Solution

The correct option is D 99x+27y+30=0
Given equations are
3x+4y+5=0
and 12x5y7=0
a1a2+b1b2=3×12+4×(5)
=16>0
For acute angle bisector
a1x+b1y+c1a21+b21=(a2x+b2y+c2)a22+b22
3x+4y+59+16=(12x5y7)122+(5)2
3x+4y+55=(12x5y7)13
39x+52y+65=60x+25y+35
99x+27y+30=0

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