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) \[\frac{\pi }{3}\] and \[\frac{x}{-2}+\frac{y}{1}=-1\]
Correct Answer: A
Solution :
Here\[a+b=-1\]. Required line is \[\frac{x}{a}-\frac{y}{1+a}=1\] .....(i) Since line (i) passes through (4, 3) \[\therefore \] \[\frac{4}{a}-\frac{3}{1+a}=1\] \[\Rightarrow \] \[4+4a-3a=a+{{a}^{2}}\] Þ \[{{a}^{2}}=4\Rightarrow a=\pm 2\] \[\therefore \] Required lines are \[\frac{x}{2}-\frac{y}{3}=1\] and\[\frac{x}{-2}+\frac{y}{1}=1\].
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