Question

Find the coordinates of the centroid of a triangle whose vertices are $A(2\sqrt{3},\sqrt{2}),B(\sqrt{27},\sqrt{8})$ and $C(\sqrt{3},-2\sqrt{2})$.

Answer 1

As we know that the centroid of a triangle whose vertices are $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given as-

Centroid $=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Given vertices of triangle are $A(2\sqrt{3},\sqrt{2}),B(\sqrt{27},\sqrt{8})$ and $C(\sqrt{3},-2\sqrt{2})$.

Therefore, the centroid of the triangle is-

Centroid of given $△=(\frac{2\sqrt{3}+\sqrt{27}+\sqrt{3}}{3},\frac{\sqrt{2}+\sqrt{8}+(-2\sqrt{2})}{3})$

$\Rightarrow$ Centroid of given $△=(\frac{2\sqrt{3}+3\sqrt{3}+\sqrt{3}}{3},\frac{\sqrt{2}+2\sqrt{2}-2\sqrt{2}}{3})$

$\Rightarrow$ Centroid of given $△=(\frac{6\sqrt{3}}{3},\frac{\sqrt{2}}{3})$

$\Rightarrow$ Centroid of given $△=(2\sqrt{3},\frac{\sqrt{2}}{3})$

Thus the centroid of the given triangle is $(2\sqrt{3},\frac{\sqrt{2}}{3})$.

Hence the correct answer is $(2\sqrt{3},\frac{\sqrt{2}}{3})$.