The centroid of triangle $ABC$ is at the point $\left(2, 3\right)$. The coordinates of $A$ and $B$ are $\left(5, 6\right)$ and $\left(-1, 4\right)$. Find the coordinates of point $C$.
Given: $\;$ $A \left(x_1, y_1\right) = \left(5, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-1, 4\right)$, $\;$ centroid $G \left(p, q\right) = \left(2, 3\right)$
Let the coordinates of point $C = \left(x, y\right)$
Let $M$ be the midpoint of $AB$.
Then, by midpoint formula, the coordinates of point $M$ are
$M = \left(\dfrac{5 - 1}{2}, \dfrac{6 + 4}{2}\right) = \left(2, 5\right)$
$CM$ is the median of $\triangle ABC$.
The centroid $G$ divides the median $CM$ internally in the ratio $2 : 1$.
$\therefore \;$ By section formula for internal division, we have,
for the $x$ coordinate: $\;$ $2 = \dfrac{2 \times 2 + 1 \times x}{2 + 1}$
i.e. $\;$ $6 = 4 + x$ $\implies$ $x = 2$
for the $y$ coordinate: $\;$ $3 = \dfrac{2 \times 5 + 1 \times y}{2 + 1}$
i.e. $\;$ $9 = 10 + y$ $\implies$ $y = -1$
$\therefore \;$ $C \left(x, y\right) = \left(2, -1\right)$