If non-zero numbers a, b, c are in harmonic p | vedclass

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(C) Given that $a, b, c$ are in harmonic progression,we have $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.This implies $\frac{1}{c} = \frac{2}{b} - \frac

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$(1, -2)$

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(C) Given that $a, b, c$ are in harmonic progression,we have $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.This implies $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$.Substitute this into the equation of the line: $\frac{x}{a} + \frac{y}{b} + (\frac{2}{b} - \frac{1}{a}) = 0$.Rearranging the terms,we get $\frac{1}{a}(x - 1) + \frac{1}{b}(y + 2) = 0$.For this equation to hold for all $a$ and $b$,the coefficients must be zero.Thus,$x - 1 = 0 \implies x = 1$ and $y + 2 = 0 \implies y = -2$.The fixed point is $(1, -2)$.

If non-zero numbers $a, b, c$ are in harmonic progression,then the straight line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ always passes through a fixed point. Find the coordinates of that point.

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