Find the equation of the bisector of the obtuse angle between the lines `3x-4y+7=0` and `12 x+5y-2=0.`

Answer

21x+77y-101=0

Explanation

Firstly, make the constant terms `(c_(1), c_(2))` positive.
3x-4y+7 = 0
and -12x-5y+2=0
`therefore a_(1)a_(2) + b_(1)b_(2) = (3)(-12) + (-4)(-5)`
=-36+20=-16
Hence, "-" sign gives the obtuse bisector.
Therefore, the obtuse bisector is
`((3x-4y+7))/(sqrt((3)^(2) + (-4)^(2))) = ((-12x-5y+2))/(sqrt((-12)^(2) + (-5)^(2)))`
or 13(3x-4y+7) = -5(-12x-5y+2)
or 21x+77y-101 =0