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

Question

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

There are infinitely many lines that pass through a certain given point. Did you copy the question correctly? Did the question ask you to compute the equation of the Euler line of this triangle (i.e., the line that passes through the circumcenter, the centroid, and the orthocenter)? — Batominovski, Jan 01 '20 at 14:56
@Batominovski Yes, there were options present in the question, but I didn’t think they were important. I will add them, but then it simply is a matter of checking the each option — Aditya, Jan 01 '20 at 15:20
@Aqua no that’s it. If the way to solve it is by using trial and error in every single option, I will see you on the other side of life by then — Aditya, Jan 01 '20 at 15:28

score 1 · Accepted Answer · answered Jan 01 '20 at 15:33

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

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

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