Let the area of the triangle formed by a straight Line L: x + by + c = 0 with co-ordinate axes be 48

Q: Wait, why can't I just use the intercept form $\frac{x}{a} + \frac{y}{b} = 1$?

You absolutely can! However, the intercept form introduces two variables ($a$ and $b$) that aren't directly linked to the angle of the normal. The Normal Form is superior here because it directly incorporates the angle $\alpha$ and the perpendicular distance $p$, which are the exact pieces of information the problem gives you. It's about choosing the tool that fits the problem's constraints, not just the one you're most comfortable with.

Q: Why is the normal angle $45^\circ$ so significant?

Great question! The angle of the normal vector determines the slope of the line. If the normal makes $45^\circ$ with the $x$-axis, the line itself must have a slope of $-1$ (since the line is perpendicular to the normal). This forces the line to be of the form $x + y = k$. Knowing this immediately simplifies the algebra, as you don't have to deal with messy trigonometric coefficients.

Q: Do I need to worry about the sign of $c$ when calculating $c^2$?

In this specific problem, no, because we are calculating $c^2$. However, in a general context, you must be careful. The perpendicular distance $p$ is always positive, but $c$ can be negative depending on which side of the origin the line lies. Since we are squaring $c$, the sign disappears, but always keep track of the geometry to ensure your logic remains robust.

Question

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

Options:

(A) 90

(B) 93

(C) 97

(D) 83

Sigma Percentile · Accepted Answer

### The Geometry of the Normal: A Masterclass in Coordinate Geometry

Welcome, future engineers! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, elegant structure beneath. We are dealing with a straight line, $L: x + by + c = 0$, slicing through the coordinate plane.

It forms a triangle with the axes, and we are given a specific constraint about a perpendicular dropped from the origin. This is a classic JEE Advanced setup—it tests not just your ability to calculate, but your ability to choose the right perspective.

### Phase 1: The Secret Weapon—The Normal Form

When you see the phrase "perpendicular drawn from the origin," your intuition should immediately fire. Most students rush to the slope-intercept form ($y = mx + c$) or the intercept form ($\frac{x}{a} + \frac{y}{b} = 1$). While those are valid, they are often the "long way home" when the problem gives you information about the normal vector.

Instead, let us invoke the **Normal Form** of a straight line:

$$x \cos \alpha + y \sin \alpha = p$$

Here, $p$ is the perpendicular distance from the origin to the line, and $\alpha$ is the angle that this perpendicular makes with the positive $x$-axis. This equation is the "skeleton" of the line.

We are given $\alpha = 45^{\circ}$. Let us substitute this into our equation:

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

Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, our equation becomes:

$$\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$$

To make this look cleaner, let us multiply the entire equation by $\sqrt{2}$:

$$x + y = p\sqrt{2}$$

Or, in the standard form we are familiar with:

$$x + y - p\sqrt{2} = 0$$

### Phase 2: The Intercepts and the Area Constraint

Now that we have the equation of the line in terms of $p$, we need to find the area of the triangle it forms with the axes. The area of a triangle formed by a line with the coordinate axes is simply $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}|$, where the base and height are the $x$ and $y$ intercepts.

To find the $x$-intercept, we set $y = 0$ in our equation $x + y = p\sqrt{2}$, which gives us $x = p\sqrt{2}$. Similarly, setting $x = 0$ gives us the $y$-intercept, $y = p\sqrt{2}$. Our intercepts are $(p\sqrt{2}, 0)$ and $(0, p\sqrt{2})$.

Now, let us calculate the area:

$$	ext{Area} = \frac{1}{2} 	imes (p\sqrt{2}) 	imes (p\sqrt{2})$$

$$	ext{Area} = \frac{1}{2} 	imes 2p^2 = p^2$$

This is the "Aha!" moment. The area is simply $p^2$. The problem tells us the area is $48$ square units. Therefore:

$$p^2 = 48$$

We do not need to find $p$ itself. We need $p^2$ for our final answer, so let us hold onto this value like a treasure.

### Phase 3: The Algebraic Comparison

We are almost there. We have our derived equation: $x + y - p\sqrt{2} = 0$. We were given the original equation: $x + by + c = 0$.

Since both equations represent the same line, their coefficients must be proportional. Comparing the two:

1. The coefficient of $x$ is $1$ in both.
2. The coefficient of $y$ is $b$ in the given equation and $1$ in our derived equation. Thus, $b = 1$.
3. The constant term is $c$ in the given equation and $-p\sqrt{2}$ in our derived equation. Thus, $c = -p\sqrt{2}$.

### Phase 4: The Final Calculation

The problem asks for the value of $b^2 + c^2$. Let us compute this step-by-step:

$$b^2 = (1)^2 = 1$$

$$c^2 = (-p\sqrt{2})^2 = 2p^2$$

We already know that $p^2 = 48$. Substituting this into our expression for $c^2$:

$$c^2 = 2 	imes 48 = 96$$

Finally, adding them together:

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

And there it is! The **answer is 97**.

Select Answer:

Visualized Solution

Visualizing the Problem

The Normal Form of a Line

Substituting $α = 4 5^{\circ}$

Simplifying the Equation

Finding the Intercepts

Calculating the Area

Solving for $p^{2}$

Comparing the Equations

Finding $c^{2}$

Final Calculation of $b^{2} + c^{2}$

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Normal

Phase 1

Phase 2

Phase 3

Phase 4

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