JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 12)
Let the area of the triangle formed by a straight line $\mathrm{L}: x+\mathrm{b} y+\mathrm{c}=0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of $45^{\circ}$ with the positive $x$-axis, then the value of $\mathrm{b}^2+\mathrm{c}^2$ is :
90
83
93
97
Explanation
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$$\begin{aligned} & L: x+b y+c=0 \\ & \because \frac{1}{2}\left|(-c) \cdot\left(\frac{-c}{b}\right)\right|=48 \\ & \therefore\left|\frac{c^2}{b}\right|=96\quad\text{..... (i)} \end{aligned}$$
Slope of line $L=-\frac{1}{b}$
$\therefore$ Slope of line perpendicular to $L$ is $b$.
$$\begin{aligned} & \therefore b=1 \\ & \therefore c^2=96 \\ & \therefore b^2+c^2=97 \end{aligned}$$
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