JEE Main 2025 Apr 2 Shift 2, Mathematics Q19: Introduction to Straight Lines

JEE Main2025Apr 2, Shift 2Mathematics

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 $4 5^{\circ}$ with the positive $x$ -axis, then the value of $b^{2} + c^{2}$ is:

A
$90$
B
$93$
C
$97$
D
$83$

Solution

Intercepts of $x + b y + c = 0$ : $x$ -intercept $= - c$ and $y$ -intercept $= - c / b$ . Area of the triangle with the axes:

$\frac{1}{2} (- c) (- \frac{c}{b}) = \frac{1}{2} \frac{c ^{2}}{∣ b ∣} = 48 \Rightarrow \frac{c ^{2}}{∣ b ∣} = 96 (⋆)$ .

The normal from the origin makes $4 5^{\circ}$ with the positive $x$ -axis. The slope of the normal is $tan 4 5^{\circ} = 1$ , while the slope of the line $L$ is $- \frac{1}{b}$ . Perpendicularity gives $1 \cdot (- \frac{1}{b}) = - 1 \Rightarrow b = 1$ .

From $(⋆)$ : $c^{2} = 96$ .

$∴ b^{2} + c^{2} = 1 + 96 = 97$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q19: Introduction to Straight Lines

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