JEE Main 2026 — 4 April, Morning Shift — Question 34

Finding Coordinates of A and B The line $L_1$ is $x = -3$. Point A: Intersection of $x = -3$ and $x - y = 0$.

$$-3 - y = 0 \implies y = -3 \implies A = (-3, -3)$$

Point B: Intersection of $x = -3$ and $3x + y = 0$.

$$3(-3) + y = 0 \implies y = 9 \implies B = (-3, 9)$$

Identifying the Obtuse Angle Bisector For lines $L_2: x - y = 0$ and $L_3: 3x + y = 0$, the bisectors are:

$$\frac{x - y}{\sqrt{1^2 + (-1)^2}} = \pm \frac{3x + y}{\sqrt{3^2 + 1^2}} \implies \frac{x - y}{\sqrt{2}} = \pm \frac{3x + y}{\sqrt{10}}$$ $$\sqrt{5}(x - y) = \pm(3x + y)$$

To find the obtuse bisector, we check $a_1a_2 + b_1b_2$:

$$(1)(3) + (-1)(1) = 3 - 1 = 2$$

Since $a_1a_2 + b_1b_2 > 0$, the positive sign gives the \textbf{obtuse} bisector:

$$\sqrt{5}x - \sqrt{5}y = 3x + y \implies (\sqrt{5} - 3)x = (\sqrt{5} + 1)y$$

Finding Point C Point $C$ lies on $L_1$ ($x = -3$):

$$(\sqrt{5} - 3)(-3) = (\sqrt{5} + 1)y_C \implies y_C = \frac{9 - 3\sqrt{5}}{\sqrt{5} + 1}$$

Calculating the Ratio $BC^2 : AC^2$ Since $A, B, C$ are collinear on the vertical line $x = -3$, the ratio of distances depends only on $y$-coordinates:

$$\frac{BC^2}{AC^2} = \left( \frac{y_B - y_C}{y_A - y_C} \right)^2$$

Calculating the differences:

$$y_B - y_C = 9 - \frac{9 - 3\sqrt{5}}{\sqrt{5} + 1} = \frac{9\sqrt{5} + 9 - 9 + 3\sqrt{5}}{\sqrt{5} + 1} = \frac{12\sqrt{5}}{\sqrt{5} + 1}$$ $$y_A - y_C = -3 - \frac{9 - 3\sqrt{5}}{\sqrt{5} + 1} = \frac{-3\sqrt{5} - 3 - 9 + 3\sqrt{5}}{\sqrt{5} + 1} = \frac{-12}{\sqrt{5} + 1}$$

The ratio is:

$$\frac{BC^2}{AC^2} = \left( \frac{12\sqrt{5}}{-12} \right)^2 = (-\sqrt{5})^2 = 5$$

Final Answer: $BC^2 : AC^2 = 5 : 1$