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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Straight Lines

Let the area of the triangle formed by a straight Line L: x+by+c=0...

Let the area of the triangle formed by a straight Line L: x+by+c=0\mathrm{x}+\mathrm{by}+\mathrm{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∘45^{\circ} with the positive x -axis, then the value of b2+c2\mathrm{b}^2+\mathrm{c}^2 is:

Options

  1. A9090
  2. B9393
  3. C9797
  4. D8383

Correct answer

C. 9797

Step-by-step solution

x−c+y−c/b=1\frac{x}{-c}+\frac{y}{-c / b}=1∴ area of  triangle =12∣c2b∣=48∣c2b∣=96∵−c=−cb⇒b=1∴c2=96∴b2+c2=97\begin{aligned} & \therefore \text{ area\text{ of } triangle }=\frac{1}{2}\left|\frac{c^2}{b}\right|=48 \\ & \left|\frac{c^2}{b}\right|=96 \\ & \because-c=-\frac{c}{b} \\ & \Rightarrow b=1 \quad \therefore c^2=96 \\ & \therefore b^2+c^2=97\end{aligned}$$
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