Q74 / 200 · Manipal (MET) 2013 · Mathematics

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
    $(K, K)$
  • Correct answer
    B
    $\left(\frac{1}{K}, \frac{1}{K}\right)$
  • C
    $(-K,-K)$
  • D
    $(K-1, K-1)$

Solution

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$ ....(i) Its intercepts on $y$ and $x$ axes are $b$ and $a$ respectively. According to the question, $\frac{1}{a}+\frac{1}{b}=$ constant $=K$ $\frac{1}{a}+\frac{1}{b}=$ constant $=K$ (say) $\therefore \quad \frac{1}{a K}+\frac{1}{b K}=1$ $\Rightarrow \quad \frac{1 / K}{a}+\frac{1 / K}{h}=1$ ....(ii) From (ii) it follows that line (i) passes tnorougn the fixed point $\left(\frac{1}{K}, \frac{1}{K}\right)$.

Hence, the correct answer is option: 'B'.

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