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. $(K,K)$
• B. $(\frac{1}{K},\frac{1}{K})$
• C. $(-K,-K)$
• D. $(K-1,K-1)$

Answer 1

Answer: (B)

$(\frac{1}{K},\frac{1}{K})$

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, we have
$\frac{1}{a}+\frac{1}{b}=$ constant $=K$ (say)
$\therefore \frac{1}{aK}+\frac{1}{bK}=1$
$\Rightarrow \frac{1/K}{a}+\frac{1/K}{b}=1$ ...(ii)
From (ii) it follows that line (i) passes through the fixed point $(\frac{1}{K},\frac{1}{K})$.