Question
The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is $$-1$$ is-
A.
$$\frac{x}{2} - \frac{y}{3} = 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = 1$$
B.
$$\frac{x}{2} - \frac{y}{3} = - 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = - 1$$
C.
$$\frac{x}{2} + \frac{y}{3} = 1\,\,{\text{and}}\,\,\frac{x}{2} + \frac{y}{1} = 1$$
D.
$$\frac{x}{2} + \frac{y}{3} = - 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = - 1$$
Answer :
$$\frac{x}{2} - \frac{y}{3} = 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = 1$$
Solution :
Let the required line be $$\frac{x}{a} + \frac{y}{b} = 1.....(1)$$
then $$a + b = - 1.....(2)$$
(1) passes through (4, 3), $$ \Rightarrow \frac{4}{a} + \frac{3}{b} = 1$$
$$ \Rightarrow 4b + 3a = ab.....(3)$$
Eliminating $$b$$ from (2) and (3), we get
$${a^2} - 4 = 0\,\, \Rightarrow a = \pm 2\,\, \Rightarrow b = - 3\,\,{\text{or}}\,1$$
$$\therefore $$ Equations of straight lines are
$$\frac{x}{2} + \frac{y}{{ - 3}} = 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = 1$$