JEE Main · 2005 · 08 May 2005 · Offline · Sequences and Series

Since $a,b,c$ are in H.P., their reciprocals are in A.P., so $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$, which rearranges to $\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=0$. Comparing this with the line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$, the point that satisfies it for all such $a,b,c$ is $x=1$, $y=-2$. Hence the line always passes through $(1,-2)$.