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

Then x-intercept = a and y-intercept = b

$\therefore a+b=1$

and ab = - 6

Putting b = $\frac{-6}{a}$ from (ii) in (i), we get

$a-\frac{6}{a}=1\iff a^2-a-6=0\iff (a-3)(a+2)=0$

$\iff a=3ora=-2$

Now, $a=3\iff b=(1-a)=(1-3)=-2$

And, $a=-2\iff b=(1-a)=(1+2)=3$

$\therefore$ the required equation is

$\frac{x}{3}+\frac{y}{-2}=1\text{or}\frac{x}{-2}+\frac{y}{3}=1$

i.e.,$2x-3y-6=0\text{or}3x-2y+6=0$