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JEE Main & Advanced Mathematics Rectangular Cartesian Coordinates Question Bank Points related to triangle (Orthocente Circumcentre Incentre), Area of some geometrical figures Collinearity

  • question_answer
    If the points \[(k,\,2-2k)\], \[(1-k,\text{ }2k)\] and \[(-k-4,\text{ }6-2k)\] be collinear, then the possible values of k are [AMU 1978; RPET 1997]

    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 \[\frac{1}{2}\].


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