Animated Solution for Mathematics - Straight Lines: A ray of light along $x+\sqrt{3}y=\sqrt{3}$ gets reflected upon reaching x-axis, the equation of the reflected ray is

Imagine you are standing in a dark room, holding a laser pointer. You aim it at a mirror lying flat on the floor—the $x$-axis. As the beam hits the surface, it bounces off, obeying the fundamental law of reflection.

In the world of coordinate geometry, this is a beautiful dance of lines and points. Today, we are going to master the reflection of a ray of light defined by the equation $x+\sqrt{3}y=\sqrt{3}$.

Every reflection begins with a collision. The ray travels through space until it meets the reflecting surface—the $x$-axis, which is defined by the equation $y=0$.

To find where our light ray strikes this mirror, we find the intersection of our line and the $x$-axis. By substituting $y=0$ into our incident ray equation:

We immediately find that $x=\sqrt{3}$. Thus, our point of incidence, $A$, is $(\sqrt{3},0)$. This is the anchor point for our reflected ray.

To find the path of the reflected ray, we use the Image Point Method. Imagine the space behind the mirror as a virtual world where the reflected ray behaves as if it originated from the virtual image of the incident ray.

We select a second point on the incident ray by setting $x=0$. Substituting this into $x+\sqrt{3}y=\sqrt{3}$, we get $\sqrt{3}y=\sqrt{3}$, which simplifies to $y=1$. Thus, we have a second point, $B(0,1)$.

The reflection of any point $(x,y)$ across the $x$-axis ($y=0$) is simply $(x,-y)$. The $x$-coordinate remains locked in place, while the $y$-coordinate flips its sign.

Applying this to our point $B(0,1)$, we find its image $B^{\prime }$ at $(0,-1)$. This point $B^{\prime }$ is our virtual source. The reflected ray must pass through both the point of incidence $A(\sqrt{3},0)$ and this virtual image point $B^{\prime }(0,-1)$.

We now define the reflected ray using the two points $A(\sqrt{3},0)$ and $B^{\prime }(0,-1)$. Using the two-point form of a line equation:

Substituting our coordinates, we obtain:

Simplifying the right side gives $\frac{-1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}$. Thus, $\frac{y}{x-\sqrt{3}}=\frac{1}{\sqrt{3}}$. Cross-multiplying yields the final equation:

In the high-pressure environment of the JEE Advanced, time is your most precious resource. When reflecting across the $x$-axis, you can simply replace $y$ with $-y$ in the original equation.

Starting with $x+\sqrt{3}y=\sqrt{3}$, replacing $y$ with $-y$ gives:

Rearranging this, we get $\sqrt{3}y=x-\sqrt{3}$. It is the exact same result, achieved in seconds. Keep this tool in your arsenal, but never forget the geometric beauty that makes it work.