Mathematics · Straight lines

JEE Main 2025 — 2 April, Evening Shift — Question 41

Let the area of the triangle formed by a straight line $L: x+b y+c=0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line $L$ makes an angle of $45^{\circ}$ with the positive $x$ axis, then the value of $b^{2}+c^{2}$ is:

Option A:
97
Correct
Option B:
90
Option C:
93
Option D:
83

Answer: A

Step-by-step solution

$L: x+b y+c=0$

$\because \frac{1}{2}\left|(-c) \cdot\left(\frac{-c}{b}\right)\right|=48$

$\therefore\left|\frac{c^{2}}{b}\right|=96 ..(i)$

Slope of line $L=-\frac{1}{b}$

$\therefore$ Slope of line perpendicular to $L$ is $b$ .

$\therefore b=1$

$\therefore c^{2}=96$

$\therefore b^{2}+c^{2}=97$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Various forms of Straight Line equations

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