Question

Find the equation of the line which passes through the point $(1,-6)$ and whose product of the intercepts on the co-ordinate axes is one

Answer 1

Let $a$ and $b$ be the $x$ and $y$-intercepts respectively Therefore, equation of line is:
$\frac{x}{a}+\frac{y}{b}=1$

Since it passes through $(1,-6)$.
$\Rightarrow \frac{1}{a}+\frac{-6}{b}=1$ ...(i)

Given, $ab=1\Rightarrow b=\frac{1}{a}$
from (i)
$\Rightarrow \frac{1}{a}-6a=1$
$\Rightarrow 1-6a^2=a$
$\Rightarrow 6a^2+a-1=0$
$\Rightarrow 6a^2+3a-2a-1=0$
$\Rightarrow 3a(2a+1)-1(2a+1)=0$
$\Rightarrow (3a-1)(2a+1)=0$
$\Rightarrow a=-\frac{1}{2},\frac{1}{3}$

When, $a=-\frac{1}{2};b=-2$
When, $a=\frac{1}{3};b=3$

Therefore, equation of lines are
$-2x-\frac{1}{2}y=1\Rightarrow 4x+y=-2$
and $3x+\frac{1}{3}y=1\Rightarrow 9x+y=3$