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

Answer 1

Answer: (D)

Given circumcenter is at $(0,0)$
Mid -point of $\left( a ^{2}+1, a ^{2}+1\right)$ and $(2 a ,-2 a )$ 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 $(2 a ,-2 a )$ 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 $(a+1)^{2} y-(a-1)^{2} x=0$