For the straight lines `4x+3y-6=0` and `5x+12 y+9=0,` find the equation of the bisector of the obtuse angle between them, bisector of the acute angle between them, and bisector of the angle which contains (1, 2)

Explanation

Equations of bisector of the angles between the given lines are: `(4x +3y-6)/(sqrt(4^(2)+3^(2))) = +-(5x +12y-9)/(sqrt(5^(2)+12^(2)))`
`"or "(4x +3y-6)/(5) = +-(5x +12y-9)/(13)`
`"or "9x -7y-41 =0 " and " 7x +9y-3 = 0`
`" If " alpha//2 " is the acute angle between the line " 4x+3y-6=0 " and the bisector " 9x-7y-41=0, "then"`
` "tan"(alpha)/(2) = |(-(4)/(3)-(9)/(7))/(1+((-4)/(3))(9)/(7))| = (11)/(3) gt 1`
`therefore (alpha)/(2) gt 45^(@) " or " alpha gt 90^(@)`
Hence, 9x-7y-41 = 0 is obtuse angle bisector.
So, 7x+9y-3=0 is acute angle bisector.
Alternatively, given lines are -4x-3y+6 = 0 and 5x+12y+9=0.
`"Here, " a_(1)a_(2) +b_(1)b_(2) = (-4)(5)+(-3)(12)lt 0`
So, acute angle bisector is
`(-4x-3y+6)/(sqrt(4^(2) + 3^(2))) = (5x+12y+9)/(sqrt(5^(2) + 12^(2)))`
`" or " (-4x-3y+6)/(5) = (5x+12y+9)/(13)`
or 7x+9-3=0
`"Now, " f_(1) (x,y) = 4x+3y-6 " and " f_(2)(x,y) = 5x+12y+9`
`f_(1) (1,2) = 4(1)+3(2)-6 gt 0`
`f_(2) (1,2) = 5(1)+12(2)+9 gt 0`
Hence, equation of the bisector of the angle containing the point (1,2) is
`(4x+3y-6)/(5) = (5x+12y+9)/(13) " or " 9x-7y-41=0`
`"Also, in original equations, "c_(1)=-6 " and "c_(2)=9.`
`"Thus, "c_(1) " and "c_(2)" have opposite signs."`
Hence, angle bisector containing origin is 7x+9y-3=0.