Q. The equation of the straight lines passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is $- 1$, is
A) $\frac{x}{2} + \frac{y}{3} = - 1\ and\frac{x}{- 2} + \frac{y}{1} = - 1$
B) $\frac{x}{2} - \frac{y}{3} = - 1\ and\frac{x}{- 2} + \frac{y}{1} = - 1$
C) $\frac{x}{2} + \frac{y}{3} = 1\ and\frac{x}{- 2} + \frac{y}{1} = 1$
D) $\frac{x}{2} - \frac{y}{3} = 1\ and\frac{x}{- 2} + \frac{y}{1} = 1$
Explanation
(d)
Let $a$ and $b$ intercepts on the coordinate axes.
$\therefore a + b = - 1\ \ \Rightarrow \ b = - (a + 1)$
$Equation\ of\ line\ is\frac{x}{a} + \frac{y}{b} = 1$
$\Rightarrow \frac{x}{a} - \frac{y}{a + 1} = 1$ …(i)
Since, this line passes through (4, 3)
$\therefore\frac{4}{a} - \frac{3}{a + 1} = 1$
$\Rightarrow \ a + 4 = a^{2} + a$
$\Rightarrow \ \ a^{2} = 4\ \Rightarrow \ a = \pm 2$
$\therefore\$Equation of line is
$\frac{x}{2} - \frac{y}{3} = 1$ or $\frac{x}{- 2} + \frac{y}{1} = 1$ [from Eq.(i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.