Animated Solution for Mathematics - Straight Lines: Let $P=(-1,0),Q=(0,0)$ and $R=(3,3\sqrt{3})$ be three points. Then the equation of the bisector of the angle $PQR$ is

Welcome, future engineers. Today, we are not just solving a coordinate geometry problem; we are learning to see the hidden symmetry in the Cartesian plane.

When you look at the points $P=(-1,0)$, $Q=(0,0)$, and $R=(3,3\sqrt{3})$, do not just see numbers. See a story. Point $Q$ is our anchor, the origin of our universe. Point $P$ is a sentinel standing on the negative x-axis, and point $R$ is a beacon shining in the first quadrant.

Our mission is to find the path that splits the angle $\angle PQR$ perfectly in two.

Before we touch a pen to paper, let us visualize. We have a line segment $QP$ lying flat on the negative x-axis. We have a line segment $QR$ shooting upward into the first quadrant.

The angle between them is obtuse, clearly greater than $90^{\circ }$. Many students rush to the standard angle bisector formula, but that is like using a sledgehammer to crack a nut. We want elegance, speed, and a deep understanding of the geometry.

To find the bisector, we need to know the orientation of these lines. Let us focus on line $QR$.

The slope $m_{QR}$ is defined as the change in $y$ over the change in $x$. With $Q(0,0)$ and $R(3,3\sqrt{3})$, we calculate:

Now, ask yourself: what angle $\theta$ has a tangent of $\sqrt{3}$? If you recall your trigonometric values, you know immediately that $\tan (60^{\circ })=\sqrt{3}$. So, line $QR$ makes an angle of $60^{\circ }$ with the positive x-axis.

Now, consider line $QP$. It lies on the negative x-axis. Measuring from the positive x-axis, we rotate all the way around to the left, which is a rotation of $180^{\circ }$. So, our two lines are defined by angles ${\theta }_1=60^{\circ }$ and ${\theta }_2=180^{\circ }$.

Here is the secret that separates the top rankers from the rest. When you have two lines passing through the origin, the angle bisector is simply the line that makes an angle equal to the average of the two lines' angles.

Because the bisector must be equidistant from both rays, by averaging the angles, we are finding the line of symmetry. Let us calculate:

This is the angle our bisector makes with the positive x-axis. It is beautiful, isn't it? No complex distance formulas, no square roots of sums of squares, just pure geometric intuition.

Now that we have the angle of the bisector, we need its equation. We know the slope $m=\tan (120^{\circ })$.

Using the identity $\tan (180^{\circ }-\theta )=-\tan (\theta )$, we find:

We have a line passing through the origin $(0,0)$ with a slope of $-\sqrt{3}$. Using the point-slope form $y-y_1=m(x-x_1)$, we get $y-0=-\sqrt{3}(x-0)$, which simplifies to $y=-\sqrt{3}x$.

Rearranging this, we arrive at the final result:

You have successfully navigated the geometry, utilized the power of trigonometric slopes, and arrived at the solution with precision. Remember, in JEE Advanced, the most elegant path is often the one that relies on the fundamental properties of the shapes themselves.