Let a point A lie between the parallel lines L1 and L2 such that its distances from L1 and L2 are 6

Q: Wait, why do we place $A$ at $(0, 3)$? Can't we place it anywhere?

That is a fantastic question! You absolutely can place $A$ anywhere, but coordinate geometry is all about choosing a frame of reference that makes the algebra collapse into something manageable. By placing $A$ on the y-axis, we ensure that its x-coordinate is zero, which removes a variable from our distance formulas. It is not a restriction; it is a strategic choice to save you time and reduce the chance of calculation errors.

Q: Why do we reject the negative root when solving for $x_C^2$?

Remember that $x_C$ represents a coordinate on the x-axis, which is a real number. When we square a real number, the result must be non-negative ($x_C^2 \ge 0$). In our quadratic equation, we found roots of $75$ and $-9$. Since a squared real number cannot be negative, $-9$ is physically impossible in this context. Always keep the physical reality of your variables in mind!

Q: Is there a way to solve this without using coordinate geometry?

Yes, there is! You could use rotation or projection methods, but those often require complex trigonometric identities or geometric constructions that are harder to visualize under exam pressure. Coordinate geometry is the 'Swiss Army Knife' of JEE problems—it might take a bit of algebra, but it is a reliable, systematic path to the correct answer every single time.

Question

Let a point A lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$, respectively, is :

Options:

(A) $15\sqrt{6}$

(B) $21\sqrt{3}$

(C) $12\sqrt{2}$

(D) 27

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Solving the Equilateral Triangle Problem

Welcome, fellow traveler on the JEE journey. Today, we are going to dissect a problem that, at first glance, feels like a spatial puzzle.

We have two parallel lines, $L_1$ and $L_2$, and a point $A$ floating between them. We need to construct an equilateral triangle $ABC$ with $B$ on $L_1$ and $C$ on $L_2$.

It sounds simple, but the beauty lies in how we translate this physical reality into the language of mathematics.

### Phase 1: The Coordinate Setup

Imagine you are standing on a vast, flat plane. We have two parallel lines, $L_1$ and $L_2$.

We know the distance from $A$ to $L_1$ is $6$ and to $L_2$ is $3$. This means the total distance between the lines is $6 + 3 = 9$.

To make our lives easier, let us define our coordinate system. Let $L_2$ be the x-axis, so its equation is $y = 0$. Consequently, $L_1$ becomes the line $y = 9$.

Now, where do we place point $A$? Let us place it on the y-axis at $(0, 3)$. This choice is strategic; it centers our problem and simplifies our distance calculations significantly.

### Phase 2: The Equilateral Constraint

We need to find the coordinates of $B$ and $C$. Since $B$ lies on $L_1$, its coordinates are $(x_B, 9)$. Since $C$ lies on $L_2$, its coordinates are $(x_C, 0)$.

The core of this problem is the equilateral triangle $ABC$. In an equilateral triangle, all sides are equal, which means the squares of the sides must also be equal: $AB^2 = AC^2 = BC^2$. This is our golden key.

Let us calculate these squares using the distance formula. For $AB^2$, we have:

$$AB^2 = (x_B - 0)^2 + (9 - 3)^2 = x_B^2 + 36$$

For $AC^2$, we have:

$$AC^2 = (x_C - 0)^2 + (0 - 3)^2 = x_C^2 + 9$$

And for the third side, $BC^2$:

$$BC^2 = (x_B - x_C)^2 + (9 - 0)^2 = (x_B - x_C)^2 + 81$$

### Phase 3: The Algebraic Dance

Now, we equate these expressions. First, $AB^2 = AC^2$ gives us:

$$x_B^2 + 36 = x_C^2 + 9 \ \Rightarrow x_B^2 = x_C^2 - 27$$

This is a powerful relation. It allows us to express $x_B^2$ in terms of $x_C^2$. Next, we equate $AC^2 = BC^2$:

$$x_C^2 + 9 = (x_B - x_C)^2 + 81$$

Expanding the right side, we get:

$$x_C^2 + 9 = x_B^2 + x_C^2 - 2x_B x_C + 81$$

Notice the $x_C^2$ terms on both sides? They cancel out, leaving us with a much simpler equation:

$$9 = x_B^2 - 2x_B x_C + 81$$

Now, substitute our earlier relation $x_B^2 = x_C^2 - 27$ into this equation:

$$9 = (x_C^2 - 27) - 2x_B x_C + 81$$

Rearranging to isolate the product term $2x_B x_C$, we find:

$$2x_B x_C = x_C^2 + 45$$

### Phase 4: The Final Calculation

To eliminate $x_B$ completely, we square both sides:

$$4x_B^2 x_C^2 = (x_C^2 + 45)^2$$

Substituting $x_B^2 = x_C^2 - 27$ again, we get a quartic equation in terms of $x_C^2$:

$$4(x_C^2 - 27)x_C^2 = (x_C^2 + 45)^2$$

Expanding both sides leads us to:

$$4x_C^4 - 108x_C^2 = x_C^4 + 90x_C^2 + 2025$$

$$3x_C^4 - 198x_C^2 - 2025 = 0$$

Dividing by $3$, we get the quadratic form $u^2 - 66u - 675 = 0$, where $u = x_C^2$. Factoring this, we find $(u - 75)(u + 9) = 0$. Since $x_C^2$ cannot be negative, we have $x_C^2 = 75$.

Finally, the side length squared $a^2 = AC^2 = x_C^2 + 9 = 75 + 9 = 84$. The area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$. Plugging in our value:

$$	ext{Area} = \frac{\sqrt{3}}{4} 	imes 84 = 21\sqrt{3}$$

And there you have it! Through the systematic application of coordinate geometry, we have tamed the problem. Remember, every complex equation is just a story waiting to be told. Keep practicing, and keep falling in love with the process.

Select Answer:

Visualized Solution

Visualizing the Setup

Setting the Coordinate System

Coordinates of $B$ and $C$

Distance $A B^{2}$ and $A C^{2}$

Distance $B C^{2}$

Equating $A B^{2}$ and $A C^{2}$

Equating $A C^{2}$ and $B C^{2}$

Substituting $x_{B}^{2}$

Squaring the Relation

Forming the Quartic Equation

Solving for $x_{C}^{2}$

Calculating the Area

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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