If `x + y = 0` is the angle bisector of the angle containing the point (1,0), for the line `3x + 4y + b = 0; 4x+3y + b =0, 4x + 3y-b = 0` then find the range of b

Explanation

Given liens are
`3x+4y+b=0 " " (1)`
`"and "4x+3y-b=0 " " (2)`
Equation of bisectors of lines (1), and (2) are
`(3x+4y+b)=+-(4x+3y-b)`
For bisector x+y=0, we have to choose negatie sign.
Thus, bisector x+y=0, goes through the region where lines 3x+4y+b and 4x+3y-b have opposite signs.
For bisector of the angle containing the point (1,0), we must have
`(3(1) +4(0)+b)(4(1)+3(0)-b) lt 0`
`rArr (3+b)(4-b) lt 0`
`rArr b gt 4 or b lt -3`