JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

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 :

A
$5 : 1$
B
$1 : 5$
C
$2 : 3$
D
$3 : 2$

Solution

Intersect $L_{1}$ with $L_{2}$ : $x = - 3, y = - 3 \Rightarrow A = (- 3, - 3)$ . Intersect $L_{1}$ with $L_{3}$ : $x = - 3, 3 (- 3) + y = 0 \Rightarrow B = (- 3, 9)$ .

The lines $L_{2}$ and $L_{3}$ both pass through the origin $O = (0, 0)$ . The angle bisector from $O$ between $L_{2}$ and $L_{3}$ meets $L_{1}$ at some point $C$ . Since the bisector from a vertex divides the opposite side in the ratio of adjacent sides emanating from that vertex,

$\frac{B C}{A C} = \frac{O B}{O A} .$

Compute $O A = 9 + 9 = 32$ and $O B = 9 + 81 = 310$ .

$∴ \frac{B C ^{2}}{A C ^{2}} = \frac{O B ^{2}}{O A ^{2}} = \frac{90}{18} = 5.$

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