Animated Solution for Mathematics - Straight Lines: Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $PQR$ is an equilateral triangle. Then the area of $\Delta PQR$ is :

Unlocking the Equilateral Triangle

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden symmetry of an equilateral triangle dancing on a coordinate plane.

Imagine you are standing on the Cartesian grid. You see a point $R$ at $(3,7)$ and a line $x+y=5$ stretching across the plane. We are tasked with finding the area of an equilateral triangle $PQR$ where $P$ and $Q$ are anchored to that line.

It sounds simple, but the beauty lies in how we bridge the gap between the point and the line.

Visualizing the Setup

First, let us ground ourselves. We have a point $R(3,7)$ and a line $L:x+y=5$. If you were to sketch this, you would see the line sloping downwards with a slope of $-1$.

The point $R$ sits comfortably above this line. The triangle $PQR$ is equilateral, meaning all its sides are equal and all its angles are $60^{\circ }$.

The base $PQ$ lies on the line $L$. This is our crucial insight: the altitude of the triangle from vertex $R$ to the base $PQ$ is simply the perpendicular distance from the point $R$ to the line $L$.

The Altitude as a Bridge

To find the area, we don't need to hunt for the coordinates of $P$ and $Q$ individually. That would be a long, winding road.

Instead, let us use the power of the perpendicular distance formula. The distance $h$ from a point $(x_1,y_1)$ to a line $ax+by+c=0$ is given by:

For our line $x+y=5$, we rewrite it in standard form as $x+y-5=0$. Here, $a=1$, $b=1$, and $c=-5$.

Plugging in our point $R(3,7)$, where $x_1=3$ and $y_1=7$, we get:

This $h$ is the altitude of our triangle. It is the backbone of our calculation.

The Geometric Shortcut

Now, we need the area. You likely know the area of an equilateral triangle with side $a$ is $A=\frac{\sqrt{3}}{4}{a}^2$. But we have $h$, not $a$.

In an equilateral triangle, the altitude $h$ and side $a$ are related by $h=\frac{\sqrt{3}}{2}a$. Rearranging for $a$, we get $a=\frac{2h}{\sqrt{3}}$.

Substituting this back into the area formula gives us a direct path:

This is the elegant shortcut we were looking for! We don't need to find the side length $a$ explicitly; we just need the square of the altitude.

We have $h=\frac{5}{\sqrt{2}}$. Squaring this gives:

Now, we simply divide by $\sqrt{3}$ to find the area:

And there it is. The area of our triangle is $\frac{25}{2\sqrt{3}}$.

It is a clean, precise result born from the harmony of geometry and algebra. Remember, in JEE Advanced, the most elegant path is often the one that uses the properties of the shape itself rather than brute-force coordinate bashing.