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 \[\frac{x}{2}+\frac{y}{3}=1\]
D) \[\frac{x}{2}-\frac{y}{3}=1\] and \[\frac{x}{-2}+\frac{y}{1}=1\]
Correct Answer: D
Solution :
[d] Let a and b the intercepts on the coordinate axes. Given a+b=-1 \[\Rightarrow b=-a-1=-(a+1)\] The equation of the line is \[x/a+y/b=1\] \[\Rightarrow \frac{x}{a}-\frac{y}{a+1}=1\] ...(i) Since this line passes through (4, 3), \[\frac{4}{a}-\frac{3}{a+1}=1\] \[\Rightarrow \frac{4a+4-3a}{a(a+1)}=1\] \[\Rightarrow a+4={{a}^{2}}+a\] \[\Rightarrow {{a}^{2}}=4\Rightarrow a=\pm 2\] Therefore, the equation of the line [from (i)] is \[\frac{x}{2}-\frac{y}{3}=1\] and \[\frac{x}{-2}+\frac{y}{1}=1\]
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