Question

Equations of th bisectors of the lines 3x-4y+7=0 and 12x+5y-2=0 are given by

Answer 1

concept :- if two lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 where c₁>0 and c₂>0
Then, equations of bisector of lines are given by
(a₁x + b₁y + c₁)√(a₁² + b₁²) = ± (a₂x + b₂y + c₂)/√(a₂² + b₂²)

Here given lines are : 3x - 4y + 7 = 0 and 12x + 5y - 2 = 0 or , -12x - 5y + 2 = 0
Now, equation of bisector of lines
(3x - 4y + 7)/√(3² + 4²) = ±(-12x - 5y + 2)/√(12² + 5²)
⇒(3x - 4y + 7)/5 = ± (-12x - 5y + 2)/13
⇒13(3x - 4y + 7) = ± 5(-12x - 5y + 2)
⇒39x - 52y + 91 = ±(-60x - 25y + 10)

∴ equations are : 39x - 52y + 91 = -60x - 25y + 10
⇒ 99x - 27y + 81 = 0
⇒11x - 3y + 9 = 0
And 39x -52y + 91 = 60x + 25y - 10
⇒21x + 77y -101 = 0