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.
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