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
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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`
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