JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

Let $A (1, 0)$ , $B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

A
$8$
B
$5$
C
$13$
D
$10$

Solution

By the angle bisector theorem, the bisector from $B$ meets $A C$ at $D$ with $\frac{A D}{D C} = \frac{B A}{B C}$ .

$B A = (2 - 1)^{2} + (- 1 - 0)^{2} = 2$ .

$B C = (\frac{7}{3} - 2)^{2} + (\frac{4}{3} + 1)^{2} = \frac{1}{9} + \frac{49}{9} = \frac{5 2}{3}$ .

So $\frac{A D}{D C} = \frac{2}{5 2 /3} = \frac{3}{5}$ .

$D$ divides $A C$ in ratio $3 : 5$ (internally) starting from $A$ :

$D = (\frac{5 \cdot 1 + 3 \cdot 7/3}{8}, \frac{5 \cdot 0 + 3 \cdot 4/3}{8}) = (\frac{3}{2}, \frac{1}{2})$ .

Slope of $B D = \frac{1/2 - ( - 1 )}{3/2 - 2} = \frac{3/2}{- 1/2} = - 3$ .

Line through $B (2, - 1)$ with slope $- 3$ : $y + 1 = - 3 (x - 2) \Rightarrow 3 x + y = 5$ .

Hence $α = 3$ , $β = 1$ , $α^{2} + β^{2} = 10$ .

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