The circumcentre of a triangle lies at the or | vedclass

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Question

Circumcentre $=(0,0)$

Centroid $ = \left( {\frac{{{{\left( {a + 1} \right)}^2}}}{2},\frac{{{{\left( {a - 1} \right)}^2}}}{2}} \right)$

We k

Vedclass · Accepted Answer

$(a - 1)^2x - (a + 1)^2y\, = 0$

Vedclass · Answer

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

${\left( {a - 1} \right)^2}x - {\left( {a + 1} \right)^2}y = 0$

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

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