Maths Q10 – JEE-MAIN 2025

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 :

Clue 1 of 3: Geometric Properties of a Line

Recall how to find the intercepts of a line and the area of a triangle formed by a line with the coordinate axes.

🧠 Full Solution Path

Step 1: Use Area Information✦ Active

The given line is $x + b y + c = 0$ . The $x$ -intercept is $- c$ (by setting $y = 0$ ) and the $y$ -intercept is $- \frac{c}{b}$ (by setting $x = 0$ ). The area of the triangle formed by the line with the coordinate axes is given by $\frac{1}{2} | x_{intercept} \cdot y_{intercept} |$ .

Area = \frac{1}{2} | (- c) \cdot (- \frac{c}{b}) | = \frac{1}{2} | \frac{c^{2}}{b} |

Given that the area is $48$ square units:

\frac{1}{2} | \frac{c^{2}}{b} | = 48 ⟹ | \frac{c^{2}}{b} | = 96 ⟹ c^{2} = 96 | b | (Equation 1)

Step 2: Use Perpendicular Information○ Expand

The perpendicular drawn from the origin to the line L makes an angle of $45^{\circ}$ with the positive $x$ -axis. The normal form of a straight line is $x \cos α + y \sin α = p$ , where $α$ is the angle the normal makes with the $x$ -axis and $p$ is the perpendicular distance from the origin.

Given $α = 45^{\circ}$ , the equation of the line is:

x \cos 45^{\circ} + y \sin 45^{\circ} = p

x (\frac{1}{\sqrt{2}}) + y (\frac{1}{\sqrt{2}}) = p

x + y = p \sqrt{2}

Rearranging to match the given form $x + b y + c = 0$ :

x + y - p \sqrt{2} = 0 (Equation 2)

💡 Teacher's Secret Hint

Remember that $p$ (perpendicular distance) is always positive.

Step 3: Compare and Calculate○ Expand

Comparing Equation 2 ( $x + y - p \sqrt{2} = 0$ ) with the given line equation ( $x + b y + c = 0$ ):

We get $b = 1$ and $c = - p \sqrt{2}$ .

Substitute $b = 1$ into Equation 1 ( $c^{2} = 96 | b |$ ):

c^{2} = 96 | 1 |

c^{2} = 96

Now, we need to find the value of $b^{2} + c^{2}$ :

b^{2} + c^{2} = (1)^{2} + 96

b^{2} + c^{2} = 1 + 96

b^{2} + c^{2} = 97

💡 Teacher's Secret Hint

Ensure consistent signs when comparing coefficients, especially for the constant term.

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Maths Question 10 – JEE-MAIN 2025

Clue 1 of 3: Geometric Properties of a Line

🧠 Full Solution Path