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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
$y- 2ax\, = 0$
$y- (a^2 + 1)x\, = 0$
$y+ x\, = 0$
$(a - 1)^2x - (a + 1)^2y\, = 0$
Solution
Circumcentre $=(0,0)$
Centroid $ = \left( {\frac{{{{\left( {a + 1} \right)}^2}}}{2},\frac{{{{\left( {a – 1} \right)}^2}}}{2}} \right)$
We know the circumcenter $(O)$,
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{{\left( {a + 1} \right)}^2}}}{2},\frac{{3{{\left( {a – 1} \right)}^2}}}{2}$
Now, these coordinates satisfies eqn given in option $(d)$
Hence, required eqn of line is
${\left( {a – 1} \right)^2}x – {\left( {a + 1} \right)^2}y = 0$
