Question

The $x$-intercept of angle bisector of angle between the lines $2x+y-2=0$ and $2x+4y+7=0$ which contains the fixed point on the family of lines $(2\cos \alpha +3\sin \alpha )x+(3\cos \alpha -5\sin \alpha )y=5\cos \alpha -2\sin \alpha$ for different values of $\alpha$, is equal to

• A. $\frac{11}{2}$
• B. $-\frac{11}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{11}{2}$

Family of lines $\cos \alpha (2x+3y-5)+\sin \alpha (3x-5y+2)=0$ will pass through fixed point i.e., point of intersection of $2x+3y-5=0$ and $3x-5y+2=0$ which is $(1,1)$

$\because 2\cdot 1+1-2>0$ and $2\cdot 1+4\cdot 1+7>0$
$\therefore$ Angle bisector
$\frac{2x+y-2}{\sqrt{5}}=+\left(\frac{2x+4y+7}{2\sqrt{5}}\right)$ will contain the point $(1,1)$
$\Rightarrow 4x+2y-4=2x+4y+7$
$\Rightarrow 2x-2y-11=0$