Let $A(4,2)$,$B(6,5)$,and $C(1,4)$ be the vertices of $\Delta ABC$. The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.

(N/A) The median $AD$ of the triangle divides the side $BC$ into two equal parts.
Therefore,$D$ is the mid-point of side $BC$.
The coordinates of the mid-point of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Here,$B = (6, 5)$ and $C = (1, 4)$.
Coordinates of $D = \left(\frac{6+1}{2}, \frac{5+4}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right)$ or $(3.5, 4.5)$.