The directions of the centroid are just the normal of the directions of the vertices. So to track down the x facilitate of the orthocenter, include the three vertex x organizes and gap by three. Rehash for the y organize.

Assume $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right),C\left( {{x}_{3}},{{y}_{3}} \right)$ be the vertices of triangle ABC.

Then, let $D(-2,3)$, $E(4,-3)$ and $F(4,5)$ be the mid-points of sides BC, CA and AB respectively.

As D is the center of BC

$\frac{{{x}_{2}}+{{x}_{3}}}{2}=-2$and $\frac{{{y}_{2}}+{{y}_{3}}}{2}=3$

${{x}_{2}}+{{x}_{3}}=-4$and ${{y}_{2}}+{{y}_{3}}=6$……(1)

Similarly, E and F are the centers of AC and AB

$\frac{{{x}_{1}}+{{x}_{3}}}{2}=4$and $\frac{{{y}_{1}}+{{y}_{2}}}{2}=-3$

${{x}_{1}}+{{x}_{3}}=8$and ${{y}_{1}}+{{y}_{3}}=-6$…….. (2)

And,

$\frac{{{x}_{1}}+{{x}_{2}}}{2}=4$and $\frac{{{y}_{1}}+{{y}_{2}}}{2}=5$

${{x}_{1}}+{{x}_{2}}=8$and ${{y}_{1}}+{{y}_{2}}=10$…… (3)

From (1), (2) and (3), we have

${{x}_{2}}+{{x}_{3}}+{{x}_{1}}+{{x}_{3}}+{{x}_{1}}+{{x}_{2}}=-4+8+8$and

${{y}_{2}}+{{y}_{3}}+{{y}_{1}}+{{y}_{3}}+{{y}_{1}}+{{y}_{2}}=6-6+10$

$2\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)=12$and $2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)=10$

${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=6$and ${{y}_{1}}+{{y}_{2}}+{{y}_{3}}=5$…….. (4)

Form (1) and (4), we get

${{x}_{1}}-4=6$and ${{y}_{1}}+6=5$

${{x}_{1}}=10$, ${{y}_{1}}=-1$

So, the coordinates of A are $(10,-1)$

From (2) and (4), we get

${{x}_{2}}+8=6$and ${{y}_{2}}-6=5$

${{x}_{2}}=-2$and $\Rightarrow {{y}_{2}}=11$

Thus, the coordinates of ‘B’ are $(-2,11)$

From (3) and (4), we get

${{x}_{3}}+8=6$and ${{y}_{3}}+10=5$

${{x}_{3}}=-2$and $\Rightarrow {{y}_{3}}=-5$

Thus, the coordinates of ‘C’ are $(-2,-5)$

Hence, the vertices of triangle ABC are $A(10,-1)$, $B(-2,11)$ and $C(-2,-5)$.

Therefore, the coordinates of the centroid of triangle ABC are

$\left( \frac{10-2-2}{3},\frac{-1+11-5}{3} \right)=\left( 2,\frac{5}{3} \right)$