Question

The equation(s) of the line(s) which cut-off intercepts on the axes whose sum and product are 1 and -6 respectively, is/are

• A. 2x-3y-6 = 0
• B. 2x-3y = -6
• C. -3x+2y = 6
• D. 3x - 2y = 6

Answer 1

Answer: (A), (C)

The correct options are
A 2x-3y-6 = 0
C -3x+2y = 6

Let the equation of the line be
$\frac{x}{a}+\frac{y}{b}=1$
Given that a+b = 1 and ab = -6$\mathrm{We}\mathrm{know}\mathrm{that}{\left(a-b\right)}^2={\left(a+b\right)}^2-4\mathrm{ab}\Rightarrow {\left(a-b\right)}^2=1-4\left(-6\right)=25\Rightarrow a-b=\pm 5\mathrm{Case}I.\mathrm{If}a-b=5\mathrm{and}a+b=1\Rightarrow a=3\mathrm{and}b=-2\mathrm{Therefore},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{will}\mathrm{be}\frac{x}{3}+\frac{y}{-2}=1\Rightarrow -2x+3y=-6\Rightarrow 2x-3y-6=0$
$\mathrm{Case}\mathrm{II}.\mathrm{If}a-b=-5\mathrm{and}a+b=1\Rightarrow a=-2,b=3\mathrm{Therefore},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{will}\mathrm{be}\frac{x}{-2}+\frac{y}{3}=1\Rightarrow -3x+2y=6$