Question

A straight line moves so that the sum of the reciprocals of its intercepts made of axes is constant. Show that the line passes through a fixed point.

Answer 1

Let $a$ and $b$ be the $x$ and $y$ intercepts respectively, then

Given : $\frac{1}{a}+\frac{1}{b}=k$

where $k$ is constant and $k\ne 0$

The equation of line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1$

Using the given relation, we can write

$\Rightarrow \frac{kx}{a}+\frac{ky}{b}=k=\frac{1}{a}+\frac{1}{b}$

$\Rightarrow \frac{1}{a}(kx-1)+\frac{1}{b}(ky-1)=0$

$\Rightarrow (ky-1)=-\frac{b}{a}(kx-1)$

$\Rightarrow (y-\frac{1}{k})=-\frac{b}{a}(x-\frac{1}{k})$

Comparing with point slope form of line :

$y-y_1=m(x-x_1)$

We get

$x_1=\frac{1}{k},y=\frac{1}{k}$

Hence, the line always passes through the point

$(x_1,y_1)=(\frac{1}{k},\frac{1}{k})$