Question

The equation of straight line belongs to the family of lines $(4x-3y)+\lambda (x+2y-11)=0$ and parallel to the acute angle bisector between the lines $x-\sqrt{3}y+11=0$ and $2\sqrt{3}x-2y+7=0$ is

• A. $x-y+7=0$
• B. $x-y+1=0$
• C. $3x+4y-25=0$
• D. $3x+4y+7=0$

Answer 1

The correct option is B $x-y+1=0$

Point of intersection of $(4x-3y)=0$ and $(x+2y-11)=0$ is $(3,4)$ .
The lines $x-\sqrt{3}y+11=0$ and $2\sqrt{3}x-2y+7=0$ are making angles ${30}^{\circ }$ and ${60}^{\circ }$ with $x-$ axis respectively.
$\therefore$ acute angle bisector will make ${45}^{\circ }$ with the $x-$axis,
So required equation is $y-4=x-3$
$\Rightarrow x-y+1=0$