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If $k=\frac{a+b}{a b}$ is a non-zero constant then the point which lies on the straight line $\frac{x}{a}+\frac{y}{b}=1$ is
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The correct answer is:
$\left(\frac{1}{k}, \frac{1}{k}\right)$
$k=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$
at $(k, k): \frac{k}{a}+\frac{k}{b}=k\left(\frac{1}{a}+\frac{1}{b}\right)=k^2 \neq 1$
at $\left(\frac{1}{k}, \frac{1}{k}\right): \frac{1}{k a}+\frac{1}{k b}=\frac{1}{k}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{k} \cdot k=1$
$\therefore \quad$ Required point is $\left(\frac{1}{k}, \frac{1}{k}\right)$
at $(k, k): \frac{k}{a}+\frac{k}{b}=k\left(\frac{1}{a}+\frac{1}{b}\right)=k^2 \neq 1$
at $\left(\frac{1}{k}, \frac{1}{k}\right): \frac{1}{k a}+\frac{1}{k b}=\frac{1}{k}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{k} \cdot k=1$
$\therefore \quad$ Required point is $\left(\frac{1}{k}, \frac{1}{k}\right)$
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