Question

Find the equations of the bisectors of the angle between the straight line 3x + 4y + 2 = 0 and 5x - 12 y - 6 = 0.

• A. 8x + y + 7 = 0
• B. 16x - 2y - 1 = 0
• C. x + 8y + 4 = 0
• D. both (b) and (c)

Answer 1

The correct option is D both (b) and (c)

The equation of the lines may be written as 3x + 4y + 2 = 0 and -5x + 12y + 6 = 0 in which the constant terms 2 and 6 are both positive.
The equation of the bisector of the angle in which the origin lies is $\frac{3x+4y+2}{\sqrt{3^3+4^2}}=\frac{-5x+12y+6}{\sqrt{(-5{)}^2+(12{)}^2}}\Rightarrow 16x-2y-1=0$
The equation of the other bisector is
$\frac{3x+4y+2}{\sqrt{3^2+4^2}}=\frac{-(-5x+12y+6)}{\sqrt{(-5{)}^2+(12{)}^2}}\Rightarrow x+8y+4=0$