Question

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$, respectively.

Answer 1

Let the intercepts cut by the given line on the axes be $a$ and $b$.

It is given that,
$a+b=$ $1$....(1)
and $ab=-6$....(2)

On solving equation (1) and (2), we obtain
$a=3$ and $b=-2$ or $a=-2$ and $b=3$

It is known that the equation of the line whose intercepts on the axes are $a$ and $b$ is $\frac{x}{a}+\frac{y}{b}=1$ or $bx+ay-ab=0$.

case1: $a=3$ and $b=-2$
In this case the equation of the line is $-2x+3y+6=0\Rightarrow 2x-3y=6$

case2: $a=-2$ and $b=3$
In this case the equation of line is $3x-2y+6=0\Rightarrow -3x+2y=6$

Thus, the required equation of the lines are $2x-3y=6$ and $-3x+2y=6$.