A) \[(-1,\ -2)\]
B) \[(-1,\ 2)\]
C) \[(1,\ -2)\]
D) \[(1,\ -1/2)\]
Correct Answer: C
Solution :
a, b, c are in H. P., then \[\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] .....(i) Given line is \[\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\] .....(ii) Subtracting both \[\frac{1}{a}(x-1)+\frac{1}{b}(y+2)=0\] Since \[a\ne 0,b\ne 0\] So, \[(x-1)=0\Rightarrow x=1\text{ and }(y+2)=0\Rightarrow y=-2\]. Trick: Checking from options, let\[a,\,\,b,\,\,c\]are\[\frac{1}{1},\frac{1}{2},\frac{1}{3}\]. Then \[x+2y+3=0\]will satisfy (c) option.
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