Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit

Q: Why do we place $P$ at $(0,0)$? Can't we just use pure geometry?

You absolutely could use pure geometry, but coordinate geometry is the 'cheat code' for JEE. By placing $P$ at the origin, you eliminate two variables immediately. It turns a complex geometric visualization into a straightforward algebraic manipulation. Always choose the coordinate system that makes the most points zero!

Q: Why do we square the distances? Why not just use the distance formula with square roots?

Great question! Square roots are messy, prone to sign errors, and difficult to manipulate algebraically. By working with $L^2$ instead of $L$, we keep the algebra clean and polynomial-based. Always look to square your equations when dealing with distance formulas to keep your workspace tidy.

Q: Why do we reject the solution $L^2 = 0$?

Physics and geometry demand reality. If $L^2 = 0$, then $L=0$, which means the triangle is just a single point. A triangle must have a non-zero area and side length. We always check our mathematical roots against the physical constraints of the problem. In JEE, the 'extraneous root' is a trap designed to catch those who don't think about the physical context!

Question

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then $(QR)^{2}$ is equal to

Sigma Percentile · Accepted Answer

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

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey of coordinate geometry.

Often, when you see a problem involving parallel lines and triangles, your instinct might be to reach for trigonometry or complex geometric theorems. But there is a more elegant path—a path of pure, unadulterated algebra. Let us walk through it together.

### Phase 1: The Power of the Origin

Imagine you are standing in a coordinate plane. You have two parallel lines separated by a distance of $5$ units.

We place point $P$ at the origin, $(0,0)$. This is our anchor.

Since $P$ is $1$ unit from one line, let's define that line, $L_1$, as $y = 1$. Because the total distance between the lines is $5$, the second line, $L_2$, must be at $y = -4$.

Just like that, we have transformed a vague geometric description into a precise mathematical landscape.

### Phase 2: The Equilateral Constraint

Now, we introduce the equilateral triangle $PQR$. We place $Q$ on $L_1$ and $R$ on $L_2$.

Let the side length of this triangle be $L$. The beauty of an equilateral triangle is that all sides are equal, which means $PQ^2 = PR^2 = QR^2 = L^2$.

For point $Q$ on $L_1$, its coordinates are $(x_1, 1)$. For point $R$ on $L_2$, its coordinates are $(x_2, -4)$.

Using the distance formula from the origin $(0,0)$, we get:

$$PQ^2 = x_1^2 + 1^2 = L^2 \implies x_1^2 = L^2 - 1$$

$$PR^2 = x_2^2 + (-4)^2 = L^2 \implies x_2^2 = L^2 - 16$$

These two equations are the foundation of our solution. They link the horizontal positions of $Q$ and $R$ directly to the side length $L$.

### Phase 3: The Algebraic Symphony

Now, let's look at the third side, $QR$. The distance between $Q(x_1, 1)$ and $R(x_2, -4)$ is given by:

$$QR^2 = (x_1 - x_2)^2 + (1 - (-4))^2 = L^2$$

$$(x_1 - x_2)^2 + 25 = L^2$$

If we expand the term $(x_1 - x_2)^2$, we get $x_1^2 + x_2^2 - 2x_1x_2 + 25 = L^2$. Substituting our expressions for $x_1^2$ and $x_2^2$, we get:

$$(L^2 - 1) + (L^2 - 16) - 2x_1x_2 + 25 = L^2$$

Simplifying this, we find $2L^2 - 17 - 2x_1x_2 + 25 = L^2$, which reduces to $L^2 + 8 = 2x_1x_2$.

### Phase 4: The Final Cancellation

We have one last hurdle: the $x_1x_2$ term. To eliminate it, we square both sides of our equation:

$$(L^2 + 8)^2 = (2x_1x_2)^2 = 4x_1^2x_2^2$$

Substituting our expressions for $x_1^2$ and $x_2^2$ one last time:

$$(L^2 + 8)^2 = 4(L^2 - 1)(L^2 - 16)$$

Expanding both sides carefully, we get:

$$L^4 + 16L^2 + 64 = 4(L^4 - 17L^2 + 16)$$

$$L^4 + 16L^2 + 64 = 4L^4 - 68L^2 + 64$$

Notice the $64$ on both sides? They vanish! We are left with $3L^4 - 84L^2 = 0$.

Factoring this, we get $3L^2(L^2 - 28) = 0$. Since $L$ cannot be zero, we arrive at the beautiful conclusion: **$L^2 = 28$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometric Setup

Defining Coordinates of $Q$ and $R$

Applying Distance Formula for $P Q^{2}$

Applying Distance Formula for $P R^{2}$

The Third Side: $Q R^{2}$

Expanding the Distance Equation

Substituting $x_{1}^{2}$ and $x_{2}^{2}$

Isolating the Product Term

Squaring to Eliminate Variables

Expanding Both Sides

Solving for $L^{2}$

Final Answer

The Sigma Insight: Distance of a Point from a Line

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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