Question

The equations 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. None of the above

Answer 1

The correct option is A $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

Equation of line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1a+b=1\Rightarrow b=-1-a\Rightarrow \frac{x}{a}+\frac{y}{-1-a}=1$

It passes through $(4,3)$

$\Rightarrow \frac{4}{a}+\frac{3}{-1-a}=1\Rightarrow \frac{4}{a}-\frac{3}{1+a}=1\Rightarrow \frac{4+4a-3a}{a(1+a)}=1\Rightarrow 4+a=a{+a}^2\Rightarrow a^2=4\Rightarrow a=\pm 2\Rightarrow b=-1-a=-1-(\pm 2)=-3,1$

So, the equation of lines are

$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

So, option A is correct