Question

The angular bisector of the lines $x+3y-2=0$ and $3x+y-5=0$ which always consisits the point belonging to the family of lines $(1+2\lambda )x+(\lambda -1)y+(3-6\lambda )=0$ is

• A. $2x-2y-3=0$
• B. $4x+4y+3=0$
• C. $4x+4y-3=0$
• D. None of the above

Answer 1

The correct option is A $2x-2y-3=0$

$(1+2\lambda )x+(\lambda -1)y+(3-6\lambda )=0$
$\Rightarrow (x-y+3)+\lambda (2x+y-6)=0$
$\therefore$ point of intersection of family of lines is $P\equiv (1,4)$
Given lines are $L_1:x+3y-2=0$ and
$L_2:3x+y-5=0$
$\left(L_1{)}_P\right(L_2{)}_P=(1+12-2)(3+4-7)=\left(11\right)\left(2\right)>0$
$\therefore$ required angular bisector is $\frac{L_1}{\sqrt{1^2+3^2}}=\frac{L_2}{\sqrt{3^2+1^2}}$
$\Rightarrow x+3y-2=3x+y-5$
$\Rightarrow 2x-2y-3=0$