Find the equation of the line on which the orthocenter lies
The centroid G is $$G=(\frac{a^2+2a+1}{2},\frac{a^2-2a+1}{2})$$
Since it divides O(circumcentre) and H(orthocentre) in the ratio 2:1
Let the orthocentre be (h,k)
$$h=\frac{3a^2+6a+3}{2}$$ $$k=\frac{a^2-6a+3}{2}$$ How should I find the equation of the line
Answer is $(a-1)^2x-(a+1)^2y=0$
There were options in the question, but checking each value is extremely time consuming, so there is bound to be a shorter method
You need not find the coordinates of the orthocenter exactly. Since the orthocenter, circumcenter and centroid are collinear, just find the equation of line passing through centroid and circumcenter, i.e. $$\dfrac{y}{x}=\frac{\dfrac{a^2-2a+1}{2}}{\dfrac{a^2+2a+1}{2}}$$ So the required equation of line is $$(a-1)^2x-(a+1)^2y=0$$