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)]You need to login to perform this action.
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