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

Let the intercepts on the $x$ and $y$ axes be $a$ and $b$ respectively.
The equation of the line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$.
Given:The line passes through $(4,3):\frac{4}{a}+\frac{3}{b}=1$ ... (1).Sum of intercepts is $-1:a+b=-1\Rightarrow b=-1-a$ ... (2).Substitute (2) into (1):$\frac{4}{a}+\frac{3}{-1-a}=1$$\frac{4(-1-a)+3a}{a(-1-a)}=1$$-4-4a+3a=-a-a^2$$-4-a=-a-a^2\Rightarrow a^2=4\Rightarrow a=2$ or $a=-2$.If $a=2,$ then $b=-1-2=-3$. The line is $\frac{x}{2}+\frac{y}{-3}=1\Rightarrow \frac{x}{2}-\frac{y}{3}=1$.If $a=-2,$ then $b=-1-(-2)=1$. The line is $\frac{x}{-2}+\frac{y}{1}=1$.These are exactly the lines given in option D.