Animated Solution for Mathematics - Straight Lines: If $a,b$ and $c$ are in A.P., then the straight line $ax+by+c=0$ will always pass through a fixed point whose coordinates are .........

Imagine you are standing on a coordinate plane, looking at a line defined by the equation $ax+by+c=0$. Usually, when we see such an equation, we think of $a$, $b$, and $c$ as fixed constants that define a single, static line.

But what if these coefficients were not just random numbers? What if they were bound by a hidden, rhythmic rule?

In this problem, we are told that $a$, $b$, and $c$ are in an Arithmetic Progression (A.P.). This simple constraint transforms our single line into a 'family of lines,' all dancing around a single, secret, fixed point. Let us uncover this mystery together.

To solve this, we must first translate the language of Arithmetic Progression into the language of algebra. If $a$, $b$, and $c$ are in A.P., the gap between the first and second term must be identical to the gap between the second and third term.

Mathematically, this is expressed as $b-a=c-b$. If we rearrange this, we get $2b=a+c$, or more conveniently for our purposes:

This is the 'universal truth' for our coefficients. No matter what specific values $a$, $b$, and $c$ take, as long as they are in A.P., they must satisfy this equation.

Comparing Equations

Now, look at the two equations we have on our hands. First, we have the equation of our line: $ax+by+c=0$. Second, we have our A.P. condition: $a-2b+c=0$.

To make the comparison crystal clear, let us rewrite the A.P. condition to match the structure of the line equation. We can write it as $a(1)+b(-2)+c=0$. Now, place them side-by-side:

Do you see it? The structure is identical! For the second equation to be true, the variables $x$ and $y$ in the first equation must be $1$ and $-2$, respectively.

Because the second equation is a direct consequence of the A.P. condition, it means that the point $(1,-2)$ will satisfy the line equation $ax+by+c=0$ regardless of the specific values of $a$, $b$, and $c$. This is the magic of coordinate geometry—we have found a point that is 'locked' in place, even as the line itself rotates and shifts.

Let us test this with a concrete example to make it real. Suppose we choose $a=1,b=2,c=3$. These are clearly in A.P.

Our line becomes $1x+2y+3=0$. If we plug in our fixed point $(1,-2)$, we get:

It works! Now, let us try another set: $a=2,b=3,c=4$. The line is $2x+3y+4=0$. Plugging in $(1,-2)$ gives:

Again, it works!

This problem is a classic example of how JEE Advanced tests your ability to see beyond the surface. It is not just about solving for $x$ and $y$; it is about recognizing that a linear constraint on the coefficients of a line equation is a geometric signal.

It tells you that the line is not just any line—it is a member of a concurrent family, all passing through a single, elegant pivot point. Whenever you encounter a problem where coefficients are related by a linear equation, remember this technique.

Compare the coefficients, find the fixed point, and watch the complexity of the problem melt away. You have the tools; now go forth and conquer the next challenge! The fixed point is $(1,-2)$.