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

Answer: (D)

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