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. $\frac{x}{2}-\frac{y}{3}=1$
• B. $\frac{x}{-2}+\frac{y}{1}=1$
• C. $-\frac{x}{3}+\frac{y}{2}=1$
• D. $\frac{x}{1}-\frac{y}{2}=1$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{x}{2}-\frac{y}{3}=1$
B $\frac{x}{-2}+\frac{y}{1}=1$

Let $\frac{x}{a}+\frac{y}{b}=1$ be the required straight line.......(i)
Where $a$ and $b$ are $x$ and $y$ intercepts respectively.
passes through $(4,3)$
$\Rightarrow \frac{4}{a}+\frac{3}{b}=1$......(ii)
Given $\text{sum}=-1\Rightarrow a+b=-1$
Substitute $b=-1-a$ in equation (ii), we get
$\Rightarrow \frac{4}{a}+\frac{3}{-1-a}=1$
$-4-4a+3a=-a-a^2$
$-4-a+a=-a^2$
$a^2=4$
$a=\pm \sqrt{4}$
$\therefore a=\pm 2$
Case i: When $a=2\text{then}b=-1-2=-3$
Thus, the required line is $\frac{x}{2}-\frac{y}{3}=1$
Case(ii):
When $a=-2$ then $b=-1+2=1$
Thus, the required line is $-\frac{x}{2}+y=1$