Question

If the circumcentre of the triangle lies at $(0,0)$ and centroid is middle point of $(a^2+1,a^2+1)$ and $(2a,-2a)$ then the orthocentre lies on:

• A. $(a-1{)}^{2}x-(a+1{)}^{2}y=0$
• B. $(a-1{)}^{2}x+(a+1{)}^{2}y=0$
• C. $(a-1{)}^{2}x+(a+1{)}^{2}y+56=0$
• D. $(a-1{)}^{2}x+(a+1{)}^{2}y-56=0$

Answer 1

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

Given coordinates of circumcentre is $(0,0)$.
Coordinates of centroid is $\left(\frac{a^2+1+2a}{2},\frac{a^2+1-2a}{2}\right)$
So, centroid is $\left(\frac{(a+1{)}^2}{2},\frac{(a-1{)}^2}{2}\right)$
We know that centroid, circumcentre, orthocentre lie on the same line.
Equation of line passing through centroid and circumcentre is
$y-0=\frac{(a-1{)}^2}{(a+1{)}^2}(x-0)$
$\Rightarrow (a-1{)}^{2}x-(a+1{)}^{2}y=0$