Question:
A straight line cuts intercepts from the axis of coordinates the sum of the reciprocals of which is a constant \(K\). Then it always passes through a fixed point
A straight line cuts intercepts from the axis of coordinates the sum of the reciprocals of which is a constant \(K\). Then it always passes through a fixed point
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A fixed point satisfies the equation independent of the parameters.
Updated On: Apr 23, 2026
- \((K, K)\)
- \(\left(\frac{1}{K}, \frac{1}{K}\right)\)
- \((-K, -K)\)
- \((K-1, K-1)\)
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The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ \frac{x}{a} + \frac{y}{b} = 1, \quad \frac{1}{a} + \frac{1}{b} = K \]
Step 2: Calculation / Simplification}
From \(\frac{1}{a} + \frac{1}{b} = K\): \(\frac{1/K}{a} + \frac{1/K}{b} = 1\)
Comparing with \(\frac{x}{a} + \frac{y}{b} = 1\):
\(x = \frac{1}{K}, y = \frac{1}{K}\) satisfies the equation for all \(a, b\).
Step 3: Final Answer
\[ \left(\frac{1}{K}, \frac{1}{K}\right) \]
\[ \frac{x}{a} + \frac{y}{b} = 1, \quad \frac{1}{a} + \frac{1}{b} = K \]
Step 2: Calculation / Simplification}
From \(\frac{1}{a} + \frac{1}{b} = K\): \(\frac{1/K}{a} + \frac{1/K}{b} = 1\)
Comparing with \(\frac{x}{a} + \frac{y}{b} = 1\):
\(x = \frac{1}{K}, y = \frac{1}{K}\) satisfies the equation for all \(a, b\).
Step 3: Final Answer
\[ \left(\frac{1}{K}, \frac{1}{K}\right) \]
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