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 $\frac{x}{a}+\frac{y}{b}=1$ be the equation of line.

It is given that a + b = 1 and ab = -6.

We know that $(a-b{)}^2=(a+b{)}^2-4ab$.

$\Rightarrow (a-b{)}^2=(1{)}^2-4\times -6$

= 1 + 24 = 25

$\Rightarrow$ a - b = $\pm$ 5

Solving a + b = 1 and a - b = 5, we have

a = 3 and b = -2

Solving a + b = 1 and a - b = 5, we have

a = -2 and b = 3

Thus the required equations are

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

$=-6\Rightarrow 2x-3y=6$

and $\frac{x}{-2}+\frac{y}{3}=1\Rightarrow 3x-2y$

$=-6\Rightarrow -3x+2y=6$