J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer If \[(3,\,\,3)\] is a vertex of a triangle and \[(-3,\,\,6)\] and \[(9,\,\,6)\] are the mid points of the two sides through this vertex, then the centroid of the triangle is

    A)  \[(3,\,\,7)\]          

    B)  \[(1,\,\,7)\]

    C)  \[(-3,\,\,7)\]        

    D)  \[(-1,\,\,7)\]

    Correct Answer: A

    Solution :

    Given,  \[A=(3,3),E=(-3,6)\] and \[F=(9,6)\] Let  \[B=({{x}_{1}},{{y}_{1}})\] and \[C=({{x}_{2}},{{y}_{2}})\] Then,   \[\frac{{{x}_{1}}+3}{2}=-3,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{y}_{1}}+3}{2}=6\] \[\Rightarrow \] \[{{x}_{1}}=-9,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}_{1}}=9\] and \[\frac{{{x}_{2}}+3}{2}=9,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{y}_{2}}+3}{2}=6\] \[\Rightarrow \] \[{{x}_{2}}=15,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}_{2}}=9\] Now, centroid \[=\left( \frac{-9+15+3}{3},\frac{9+9+3}{3} \right)\] \[=(3,7)\]

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