JEE Main 2026 — Straight Lines Question with Solution

The intersection point of $L_1:x=-3$ and $L_2:x-y=0$ is $A(-3,-3)$.

The intersection point of $L_1:x=-3$ and $L_3:3x+y=0$ is $B(-3,9)$.

The lines $L_2$ and $L_3$ intersect at the origin $O(0,0)$.

The distances from the origin to $A$ and $B$ are:
$OA=\sqrt{(-3{)}^2+(-3{)}^2}=3\sqrt{2}$
$OB=\sqrt{(-3{)}^2+9^2}=3\sqrt{10}$

The angle $\angle AOB$ between the segments $OA$ and $OB$ can be determined using the dot product of vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$:
$\overrightarrow{OA}\cdot \overrightarrow{OB}=(-3)(-3)+(-3)(9)=9-27=-18$

Since the dot product is negative, $\cos (\angle AOB)<0$, which means $\angle AOB$ is an obtuse angle. Therefore, the internal bisector of $\angle AOB$ is the bisector of the obtuse angle between the lines $L_2$ and $L_3$.

According to the internal angle bisector theorem in $△AOB$, the bisector of $\angle AOB$ divides the opposite side $AB$ at point $C$ in the ratio of the adjacent sides:
$\frac{AC}{BC}=\frac{OA}{OB}=\frac{3\sqrt{2}}{3\sqrt{10}}=\frac{1}{\sqrt{5}}$

Squaring both sides yields:
$\frac{A{C}^2}{B{C}^2}=\frac{1}{5}\Rightarrow \frac{B{C}^2}{A{C}^2}=5$

Thus, $B{C}^2:A{C}^2=5:1$.

Answer: $5:1$