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

Explanation

(i) For point `(0,0)4x+3y-6and 5x+12y+9` are or opposite signs.
Hence equation of the bisector of the angle between the given straight lines containing origin is given by
`(4x+3y-6)/(5)=-(5x+12y+9)/(13)`
`implies52x+39y-78=-25x-60y-45`
`implies7x+9y-3=0`
(ii) Writing the equation of the straight lines so that constant become positive. We have `-4x-3y+6=0`
`and 5x+12y+9=0`
The equations of the bisectors of the angle between given straight lines are given by
`(5x+12y+9)/(13)=+-(-4-3y+6)/(5)" "...(i)`
Here we observe that,
`aa_(1)+bb_(1)=-20-36=-56lt0`
Hence taking negative sign of (i) we get the equation of obtuse angle bisector as
`25x+60y+45=52x+39y-78`
`implies27x-21y-123=0`
`implies9x-7y-41=0`
(iii) Taking positive sing or (i) we get, the required equation of the acute angle bisector as `7x+9y-3=0`