A(4, 7) is the vertex of triangle ABC. BA= 5 3 and AC= 4 −3 . (a) Find the coordinates of B and C. (b) If M is the midpoint of the line BC, find AM.

We are given the coordinates of $A(4, 7)$ and the vector $\vec{BA} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$.
The vector $\vec{BA}$ is calculated as $A - B$.
Let $B = (x_B, y_B)$.

$$\begin{pmatrix} 4 \\ 7 \end{pmatrix} - \begin{pmatrix} x_B \\ y_B \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$

This gives us two equations:
$4 - x_B = 5 \implies x_B = 4 - 5 = -1$
$7 - y_B = 3 \implies y_B = 7 - 3 = 4$
So, the coordinates of $B$ are $(-1, 4)$.

We are given the coordinates of $A(4, 7)$ and the vector $\vec{AC} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$.
The vector $\vec{AC}$ is calculated as $C - A$.
Let $C = (x_C, y_C)$.

$$\begin{pmatrix} x_C \\ y_C \end{pmatrix} - \begin{pmatrix} 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$$

This gives us two equations:
$x_C - 4 = 4 \implies x_C = 4 + 4 = 8$
$y_C - 7 = -3 \implies y_C = 7 - 3 = 4$
So, the coordinates of $C$ are $(8, 4)$.

The coordinates are: $\boxed{B(-1, 4), C(8, 4)}$

$M$ is the midpoint of $BC$. We use the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
Using $B(-1, 4)$ and $C(8, 4)$:

$$M = \left( \frac{-1 + 8}{2}, \frac{4 + 4}{2} \right) = \left( \frac{7}{2}, \frac{8}{2} \right) = (3.5, 4)$$

So, the coordinates of $M$ are $(3.5, 4)$.

The vector $\vec{AM}$ is calculated as $M - A$.
Using $A(4, 7)$ and $M(3.5, 4)$:

$$\vec{AM} = \begin{pmatrix} 3.5 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 7 \end{pmatrix} = \begin{pmatrix} 3.5 - 4 \\ 4 - 7 \end{pmatrix} = \begin{pmatrix} -0.5 \\ -3 \end{pmatrix}$$

The vector $\vec{AM}$ is: $\boxed{\begin{pmatrix} -0.5 \\ -3 \end{pmatrix}}$