1.

Equation of the bisector of the acute angle between lines 3x+4y+5=0 and 12x -5y-7=0 is

Answer»

21x + 77y + 100=0
99x -27y +30=0
99x +27y +30=0
21x-77y -100=0

Solution :Given lines are 3x + 4y + 5=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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