Maths Q11 – JEE-MAIN 2026

Let the line

L_{1} : x + 3 = 0

intersect the lines

L_{2} : x - y = 0

and

L_{3} : 3 x + y = 0

at the points A and B, respectively. Let the bisector of the obtuse angle between the lines

L_{2}

and

L_{3}

intersect the line

L_{1}

at the point C. Then

B C^{2} : A C^{2}

is equal to:

Clue 1 of 3: Geometric Setup

Identify the points of intersection A, B, and the common intersection point of lines $L_{2}$ and $L_{3}$ . Notice that A, B, and C are collinear.

🧠 Full Solution Path

Step 1: Identify Key Points and Geometry✦ Active

The line $L_{1}$ is $x = - 3$ . Point A is the intersection of $L_{1}$ and $L_{2} : x - y = 0$ . Substituting $x = - 3$ into $L_{2}$ gives $- 3 - y = 0 ⟹ y = - 3$ . So, $A = (- 3, - 3)$ . Point B is the intersection of $L_{1}$ and $L_{3} : 3 x + y = 0$ . Substituting $x = - 3$ into $L_{3}$ gives $3 (- 3) + y = 0 ⟹ y = 9$ . So, $B = (- 3, 9)$ . The lines $L_{2}$ and $L_{3}$ intersect at the origin $O = (0, 0)$ . Points A, B, and C all lie on the line $L_{1} : x = - 3$ .

Step 2: Apply Angle Bisector Theorem○ Expand

The line segment $O C$ is the bisector of $∠ A O B$ (where $O$ is the origin). According to the angle bisector theorem, the point $C$ divides the line segment $A B$ (externally or internally) in the ratio of the lengths of the sides $O A$ and $O B$ . Thus, $\frac{A C}{B C} = \frac{O A}{O B}$ . We need to calculate the lengths $O A$ and $O B$ :

O A = \sqrt{(- 3 - 0)^{2} + (- 3 - 0)^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}

O B = \sqrt{(- 3 - 0)^{2} + (9 - 0)^{2}} = \sqrt{9 + 81} = \sqrt{90} = 3 \sqrt{10}

💡 Teacher's Secret Hint

The 'obtuse angle bisector' implies that C is the intersection of the external bisector with line $L_{1}$ , but the magnitude of the ratio $A C : B C = O A : O B$ remains valid.

Step 3: Calculate the Required Ratio○ Expand

Substitute the values of $O A$ and $O B$ into the ratio from the angle bisector theorem:

\frac{A C}{B C} = \frac{3 \sqrt{2}}{3 \sqrt{10}} = \frac{\sqrt{2}}{\sqrt{10}} = \frac{1}{\sqrt{5}}

Therefore, $\frac{B C}{A C} = \sqrt{5}$ . The required ratio is $B C^{2} : A C^{2}$ :

\frac{B C^{2}}{A C^{2}} = (\sqrt{5})^{2} = 5

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

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Geometric Setup

🧠 Full Solution Path