Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.
Equation of t...
Question
Equation of the bisector of the acute angle between lines 3x+4y+5=0 and 12x−5y−7=0 is:
A
21x+77y+100=0
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B
99x−27y+30=0
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C
99x+27y+30=0
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D
21x−77y−100=0
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Solution
The correct option is D99x+27y+30=0 Given equations are 3x+4y+5=0 and 12x−5y−7=0 ∴a1a2+b1b2=3×12+4×(−5) =16>0 ∴ For acute angle bisector a1x+b1y+c1√a21+b21=−(a2x+b2y+c2)√a22+b22 ∴3x+4y+5√9+16=−(12x−5y−7)√122+(−5)2 ⇒3x+4y+55=−(12x−5y−7)13 ⇒39x+52y+65=−60x+25y+35 ⇒99x+27y+30=0