2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points \[({{a}^{2}}+1,{{a}^{2}}+1)\]and \[(2a,-2a),a\ne 0.\]Then for any a, the orthocentre of this triangle lies on the line:     JEE Main Online Paper (Held On 19 April 2016)

    A) \[y-2ax=0\]                        

    B) \[y-({{a}^{2}}+1)x=0\]

    C) \[y+x=0\]                           

    D) \[{{(a-1)}^{2}}x-{{(a+1)}^{2}}y=0\]

    Correct Answer: D

    Solution :

    Circumcentre = (0, 0) Centroid \[=\left( \frac{{{(a+1)}^{2}}}{2},\frac{{{(a+1)}^{2}}}{2} \right)\] We know the circumcentre (O), Centroid (G) and orthocentre (H) of a triangle lie on the line joining the O and G. Also, \[\frac{HG}{GO}=\frac{2}{1}\] \[\Rightarrow \]Coordinate of orthocentre \[=\frac{3{{(a+1)}^{2}}}{2},\frac{3{{(a-1)}^{2}}}{2}\] Now, these coordinates satisfies eqn given in option (d) Hence, required eqn of line is \[{{(a-1)}^{2}}x-{{(a+1)}^{2}}y=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner