Question

The equation of bisectors of two lines $L_1$ & $L_2$ are $2x-16y-5=0$ and $64x+8y+35=0$. If the line $L_1$ passes through $(-11,4)$, the equation of acute angle bisector of $L_1$ and $L_2$ is :

• A. $2x-16y-5=0$
• B. $64x+8y+35=0$
• C. $2x+16y+5=0$
• D. $2x+16y-5=0$

Answer 1

The correct option is A $2x-16y-5=0$

According to question,

$ly>x,putthepointof(-11,4)PA=\left|\frac{64(-11)+8\times 4+35}{\sqrt{{\left(64\right)}^2+{\left(8\right)}^2}}\right|=\left|\frac{-704+67}{4\sqrt{{\left(16\right)}^2+{\left(2\right)}^2}}\right|=\left|\frac{-637}{4\sqrt{k}}\right|=\frac{159.25}{\sqrt{k}}[suppose(16{)}^2+(2{)}^2=kNow,PB=\left|\frac{-22-64-5}{\sqrt{{\left(16\right)}^2+{\left(2\right)}^2}}\right|=\left|\frac{-91}{\sqrt{k}}\right|=\frac{91}{\sqrt{k}}so,wecansay\sin \frac{y}{2}>\sin \frac{x}{2},(y>x)\therefore acuteanglebisectoris2x-16y-5=0thecorrectoptionisA.$