A) \[\frac{1}{a}+\frac{1}{b}=2\]
B) \[\frac{1}{a}-\frac{1}{b}=1\]
C) \[\frac{1}{a}-\frac{1}{b}=2\]
D) \[\frac{1}{a}+\frac{1}{b}-1\]
Correct Answer: D
Solution :
Since, the given points A (a, 0), B(0, b) and C(1, 1) are collinear. \[\therefore \] Area of \[\Delta ABC=0\] \[\Rightarrow \] \[\frac{1}{2}|a(b-1)+0(1-0)+1(0-b)|=0\] \[\Rightarrow \] \[ab-a-b=0\,\,\,\Rightarrow \,\,a+b=ab\] Dividing both sides by ab, we get. \[\frac{1}{b}+\frac{1}{a}=1\] \[\Rightarrow \] \[\frac{1}{a}+\frac{1}{b}=1\]
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