question_answer

The equation of the straight line 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) \[\frac{x}{2}-\frac{y}{3}\,=1\] and \[\frac{x}{-2}\,+\frac{y}{1}\,=1\]

Correct Answer: D

Solution :

If a and b are intercepts on the X-axis and y-axis respectively, then the equation of the line is \[\frac{x}{a}+\frac{y}{b}=1\]. Let a and b be intercepts on the coordinate axes \[\therefore \]  \[a+b=-1\] \[\Rightarrow \]               \[b=-a-1=-(a+1)\] Equation of line is \[\frac{x}{a}+\frac{y}{b}=1\] \[\Rightarrow \]               \[\frac{x}{a}-\frac{y}{a+1}=1\]                                   ?.(i) 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 \[\frac{x}{2}-\frac{y}{3}=1\]and\[\frac{x}{-2}+\frac{y}{1}=1\]       [from Eq. (i)]