Animated Solution for Mathematics - Straight Lines: If non zero numbers $a,b,c$ are in H.P., then the straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$ always passes through a fixed point. That point is

The problem presents a family of lines defined by the equation:

Here, $a$, $b$, and $c$ are parameters that vary, causing the line to shift across the plane. However, we are given a specific constraint: $a$, $b$, and $c$ are in Harmonic Progression (H.P.).

By definition, if three numbers are in H.P., their reciprocals must be in Arithmetic Progression (A.P.). Therefore, the terms $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ form an A.P.

For any three terms $X,Y,Z$ in A.P., the middle term is the arithmetic mean of the other two, expressed as $2Y=X+Z$. Applying this property to our reciprocals, we obtain:

Rearranging this equation to set it to zero, we get:

Now, we compare the constraint equation with our original line equation. We can rewrite the line equation as:

By aligning this with the constraint $\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=0$, we observe a direct correspondence between the coefficients of the reciprocals. Specifically, we can map the terms as follows:

This confirms that for any values of $a,b,c$ satisfying the H.P. condition, the equation is always satisfied by the coordinates $(1,-2)$.

The family of lines defined by the given equation, subject to the constraint that $a,b,c$ are in H.P., is concurrent. All such lines pass through the fixed point $(1,-2)$.

This result demonstrates how algebraic constraints on parameters can reveal geometric stability in a family of lines. Whenever you encounter a linear relation between parameters, look for this underlying structure to identify the point of concurrency.