If k = frac{a+b}{ab} is a non-zero constant,t | vedclass

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(D) Given $k = \frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$.We check the point $\left(\frac{1}{k}, \frac{1}{k}\right)$ in the equation $\frac{x}{a} + \

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$\left(\frac{1}{k}, \frac{1}{k}\right)$

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(D) Given $k = \frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$.We check the point $\left(\frac{1}{k}, \frac{1}{k}\right)$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$:Substitute $x = \frac{1}{k}$ and $y = \frac{1}{k}$:$\frac{1}{ka} + \frac{1}{kb} = \frac{1}{k} \left(\frac{1}{a} + \frac{1}{b}\right)$.Since $\frac{1}{a} + \frac{1}{b} = k$,we get:$\frac{1}{k} \cdot k = 1$.Thus,the point $\left(\frac{1}{k}, \frac{1}{k}\right)$ satisfies the equation of the line.

If $k = \frac{a+b}{ab}$ 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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