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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are \[x+y=5\]and\[x=4\] respectively, Then area of \[\Delta ABC\] (in sq. units) is [JEE Online 15-04-2018]

    A) 5                                 

    B) 9         

    C) 12                               

    D) 4         

    Correct Answer: B

    Solution :

    Median through C is \[x=4\] So clearly the \[x\] coordinate of \[C\] is 4. So let \[C=(4,y),\] then the midpoint of \[A(1,2)\] and \[C(2,y)\] which is \[D\] lies on the median through \[B\] by definition. Clearly, \[D=(\frac{1+4}{2},\frac{2+y}{2})\]. Now, we have, \[\frac{3+4+y}{2}=5\Rightarrow y=3\]. So, \[C=(4,3)\]. The centroid of the triangle is the intersection of the medians. It is easy to see that the medians \[x=4\] and \[x+y=5\] Intersect at \[G=(4,1)\]. The area of triangle \[\Delta ABC=3\times \Delta AGC=3\times \frac{1}{2}\times 3\times 2=9\]. (In this case, it is easy as the points \[G\] and \[C\]lie on the same vertical line \[x=4\]. So the base \[GC=2\] and the altitude from \[A\] is \[3\] units.) So the answer is option B.


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