TS EAMCET · Maths · Straight Lines
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
- A \((k, k)\)
- B \(\left(k, \frac{1}{k}\right)\)
- C \(\left(\frac{1}{k}, k\right)\)
- D \(\left(\frac{1}{k}, \frac{1}{k}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{1}{k}, \frac{1}{k}\right)\)
Step-by-step Solution
Detailed explanation
\(k=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}\) at \((k, k): \frac{k}{a}+\frac{k}{b}=k\left(\frac{1}{a}+\frac{1}{b}\right)=k^2 \neq 1\) at \(\left(\frac{1}{k}, \frac{1}{k}\right): \frac{1}{k a}+\frac{1}{k b}=\frac{1}{k}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{k} \cdot k=1\)…
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