JEE Mathematics 2025 Question with Solution

Given: The line is $x + by + c = 0$ and the area of the triangle formed with the coordinate axes is $48$ square units.

Find: The value of $b^2 + c^2$.

The intercept form of the line is

$$\frac{x}{-c} + \frac{y}{-c/b} = 1$$

So the x-intercept is $-c$ and the y-intercept is $-\frac{c}{b}$.

Hence, the area of the triangle formed by the line with the coordinate axes is

$$\text{Area} = \frac{1}{2}\left|\frac{c^2}{b}\right| = 48$$

Therefore,

$$\left|\frac{c^2}{b}\right| = 96$$

The perpendicular from the origin to the line makes an angle of $45^\circ$ with the positive x-axis. For the line $x + by + c = 0$, the normal vector is $(1,b)$, so

$$\tan 45^\circ = \frac{b}{1}$$

Thus,

$$b = 1$$

Substituting into the area relation,

$$c^2 = 96$$

Therefore,

$$b^2 + c^2 = 1 + 96 = 97$$

The correct option is C.

Given: A straight line $L: x + by + c = 0$ forms a triangle with the coordinate axes of area $48$ square units. The perpendicular from the origin to this line makes an angle of $45^\circ$ with the positive x-axis.

Find: $b^2 + c^2$.

For the line $Ax + By + C = 0$, the direction of the perpendicular from the origin is the direction of the normal vector $(A,B)$. Here,

$$A = 1, \quad B = b, \quad C = c$$

Since the perpendicular makes an angle $45^\circ$ with the positive x-axis,

$$\tan 45^\circ = \frac{b}{1}$$

So,

$$b = 1$$

Now find the intercepts of the line.

Setting $y = 0$ gives the x-intercept:

$$x + c = 0 \Rightarrow x = -c$$

Setting $x = 0$ gives the y-intercept:

$$by + c = 0 \Rightarrow y = -\frac{c}{b}$$

Therefore, the triangle formed with the axes has base $|c|$ and height $\left|\frac{c}{b}\right|$. Its area is

$$48 = \frac{1}{2} \times |c| \times \left|\frac{c}{b}\right|$$

That is,

$$48 = \frac{c^2}{2|b|}$$

Using $b = 1$,

$$48 = \frac{c^2}{2}$$

Hence,

$$c^2 = 96$$

Now,

$$b^2 + c^2 = 1^2 + 96 = 97$$

Therefore, the required value is $97$, so the correct option is C.