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 lines on the axes be a and b.
It is given that
$a + b = 1 … (1)$
$ab = -6 … (2)$
On solving equations (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 $bx+ay-ab=0$

**Case I: ** $a = 3$ and $b = -2$
In this case, the equation of the line is $-2x + 3y + 6 = 0, i.e., 2x \- 3y = 6.$

Case II: $a = -2$ and $b = 3$
In this case, the equation of the line is $3x \- 2y + 6 = 0, i.e., -3x + 2y = 6.$

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