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)

Options

A. `9x -7y -41 =0`
B. `7x + 9y -3=0`
C. `9x -7y - 3=0`
D. `7x ++ 9x -41=0`

Answer

A

Explanation

To find the bisector of the angle between the lines which constains the origin, we find wire down the equation of the given line in such a from that the constant terms in the equations of the lines are positive. The equations of the given lines are
`-4x -3y + 6 =0" ".....(i)`
and `4x+ 12y + 9 =0" ".....(ii)`
Now, the equation of the bisector of the angle between the lines which constains the origin is the bisector corresponding to the positive sign, i.e.,
`(-4x -3y + 6)/(sqrt((-4)^(2)) + (-3)^(2)) =+(5x + 12y +9)/(sqrt(5^(2) + 12^(2)))` [Taking +ive sign ]
`rArr -52x- 39y + 78 = 25x + 60y + 45`
`rArr 7x 9y -3 =0`.