Animated Solution for Mathematics - Straight Lines: If $x_1,x_2,x_3$ as well as $y_1,y_2,y_3$ are in G.P. with the same common ratio, then the points $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$

Imagine you are standing on a vast, empty coordinate plane. You have three points, $P_1,P_2$, and $P_3$, scattered somewhere in this space.

At first glance, they seem disconnected, random, and perhaps even chaotic. But the problem gives us a secret key: the $x$-coordinates and $y$-coordinates are both in a Geometric Progression (G.P.) with the same common ratio $r$.

This isn't just a sequence; it is a hidden geometric structure waiting to be unveiled. Let's embark on a journey to discover the path these points follow.

Let's start by defining our points. If the $x$-coordinates $x_1,x_2,x_3$ are in G.P., we can express them as $x_1,x_{1}r,x_1{r}^2$.

Similarly, the $y$-coordinates $y_1,y_2,y_3$ are $y_1,y_{1}r,y_1{r}^2$. Suddenly, our points $P_1,P_2,P_3$ take on a beautiful, symmetric form:

Do you see the pattern? Each point is just a scaled version of the previous one. This symmetry is the heartbeat of the problem.

A JEE Favorite

How do we prove these points lie on a straight line? We could use the area of a triangle formula, but that is a long, winding road.

Instead, let's use the slope method—a true favorite in the JEE Advanced toolkit. If the slope of the segment $P_1{P}_2$ is equal to the slope of the segment $P_2{P}_3$, then the points must be collinear.

The slope formula is defined as:

First, let's calculate the slope of $P_1{P}_2$, which we will call $m_{12}$. Substituting our G.P. coordinates, we get:

Now, watch the magic happen. We factor out $y_1$ from the numerator and $x_1$ from the denominator:

As long as $r eq 1$, we can cancel the $(r-1)$ terms. We are left with $m_{12}=\frac{y_1}{x_1}$.

Now, let's do the same for $P_2{P}_3$, which we will call $m_{23}$. Substituting the coordinates, we get:

Again, we factor out $y_{1}r$ from the numerator and $x_{1}r$ from the denominator:

The $r$ cancels, the $(r-1)$ cancels, and we are left with $m_{23}=\frac{y_1}{x_1}$.

Look at what we have achieved! The slope of the first segment is identical to the slope of the second segment. Both are equal to $\frac{y_1}{x_1}$.

Since they share the common point $P_2$ and have the same slope, they cannot form a triangle. They must lie on a single, continuous straight line.

In fact, the equation of this line is:

This reveals that this line passes directly through the origin. We have taken a seemingly complex problem and reduced it to an elegant, simple truth. This is the beauty of mathematics—finding order in the chaos.