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Equation of the bisector of the acute angle between lines 3x+4y+5=0 and 12x -5y-7=0 is |
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Answer» 21x + 77y + 100=0 and12 x-5y -7=0 Here ` a_(1) = 3, b_(1) = 4, a_(2) =12 and b_(2) = -5 ` `a_(1)a_(2) +b_(1)b_(2) = 3xx 12 + 4 xx (-5) = 16 lt 0` For acute ANGLE bisector ` (3x + 4y+5)/(sqrt(9+16)) = ((12x-5y-7))/(sqrt(12^(2)+(-5)^(2))` ` (3x+4y+5)/5 = ((12x-5y-7))/13` ` Rightarrow 39X + 52y + 65 =- 60 x + 25 y + 35` ` Rightarrow 99 x + 27 y + 30 =0` |
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