A) \[1\]
B) \[\frac{3}{2}\]
C) \[2\]
D) \[\frac{5}{2}\]
E) \[3\]
Correct Answer: A
Solution :
Let the third vertex be (a, b). \[\therefore \]Area of\[\Delta =\frac{1}{2}\left| \left| \begin{matrix} 0 & 0 & 1 \\ a & b & 1 \\ 6 & 8 & 1 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}|[8a-6b]|\] As (a, b) are integers, so we take (0, 0), (1, 1), (1, 2) At (0, 0),\[\Delta =0,\]it is not possible. At (1, 1),\[\Delta =1\] At (1, 2),\[\Delta =2\] Here, we see that minimum area is 1.
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