Question

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 :

• A. $y-2ax=0$
• B. $y(a^2+1)-x=0$
• C. $y+x=0$
• D. $(a-1{)}^{2}x-(a+1{)}^{2}y=0$

Answer 1

The correct option is D $(a-1{)}^{2}x-(a+1{)}^{2}y=0$

Given circumcenter is at $(0,0)$
Mid -point of $(a^2+1,a^2+1)$ and $(2a,-2a)$ is $(\frac{(a+1{)}^2}{2},\frac{(a-1{)}^2}{2})$
So, coordinates of centroid is $(\frac{(a+1{)}^2}{2},\frac{(a-1{)}^2}{2})$
Equation of line joining circumcenter and centroid is
$y-0=\frac{(a-1{)}^2}{(a+1{)}^2}(x-0)$
$\Rightarrow (a+1{)}^{2}y-(a-1{)}^{2}x=0$ ....(1)
The orthocenter lies on the line joining the circumcentre and the centroid.
Here, the orthocenter satisfies eqn (1).
Hence, orthocenter lies on the line $(a+1{)}^{2}y-(a-1{)}^{2}x=0$