Mathematics · Straight lines

JEE Main 2026 — 24 January, Morning Shift — Question 13

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

Option A:
$8$
Option B:
$5$
Option C:
$13$
Option D:
$10$
Correct

Answer: D

Step-by-step solution

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{2} \times 3}{5 \sqrt{2}}=\frac{3}{5}$

$\mathrm{D}=\left(\frac{12}{8}, \frac{4}{8}\right)=\left(\frac{3}{2}, \frac{1}{2}\right)$

Slope of $\mathrm{AD}=\frac{-3 / 2}{\frac{1}{2}}=-3$

$3 \mathrm{x}+\mathrm{y}=5$

$\alpha=3, \beta=1 ; \alpha^{2}+\beta^{2}=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle bisectors, concurrent lines.

Let a line L passing through the point P(1,1,1) be perpendicular to the lines x-4/4=y-1/1=z-1/1 and frac x-171=frac y-711=frac z0 . Let the…

Answer: D

Step-by-step solution

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