Question:
Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.
Solution:
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 $b x+a y-a b=0$
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.