A) \[\frac{1}{2},-1\]
B) \[1,-\frac{1}{2}\]
C) \[1,-2\]
D) \[2,-1\]
Correct Answer: A
Solution :
The points are collinear if the area of triangle formed by these three points is zero. \[\Rightarrow \,\,\frac{1}{2}[k\{2k-(6-2k)\}+(1-k)\{(6-2k)-(2-2k)\}\]\[+(-4-k)\{(2-2k)-2k\}]=0\] On simplification, we get \[k=-1\] or \[\frac{1}{2}\].You need to login to perform this action.
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