question_answer

    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)  \[\frac{x}{2}+\frac{y}{3}=-1\]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\[\frac{x}{-2}+\frac{y}{1}=1\]

Correct Answer: D

Solution :

                 Let a and b be intercepts on the co-ordinate axes. \[\therefore \]  \[a+b=-1\] \[\Rightarrow \]               \[b=-a-1=-(a+1)\] Equation of line is \[\frac{x}{a}+\frac{y}{b}=1\] \[\Rightarrow \]               \[\frac{x}{a}-\frac{y}{a+1}=1\] Since, this line passes through (4, 3). \[\therefore \]  \[\frac{4}{a}-\frac{3}{a+1}=1\Rightarrow \frac{4a+4-3a}{a(a+1)}=1\] \[\Rightarrow \]               \[a+4={{a}^{2}}+a\] \[\Rightarrow \]               \[{{a}^{2}}=4\Rightarrow a=\pm 2\] \[\therefore \]Equation of line is \[\frac{x}{2}-\frac{y}{3}=1\] or\[\frac{x}{-2}+\frac{y}{1}=1\]        [from Eq.(i)]