Question: Three students, A, B, and C, are standing on a Cartesian coordinate plane representing a school field. Their respective standing positions are plotted at A($-2, 3$), B($4, 3$), and C($1, -1$).
Calculate the total enclosed area of the triangular region $\Delta ABC$ formed by the students.
1. **State the problem:** We need to find the area of triangle $\Delta ABC$ with vertices at $A(-2, 3)$, $B(4, 3)$, and $C(1, -1)$.
2. **Formula for area of a triangle given coordinates:** The area $A$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:
$$
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$
3. **Assign coordinates:**
- $x_1 = -2$, $y_1 = 3$
- $x_2 = 4$, $y_2 = 3$
- $x_3 = 1$, $y_3 = -1$
4. **Substitute values into the formula:**
$$
A = \frac{1}{2} \left| (-2)(3 - (-1)) + 4((-1) - 3) + 1(3 - 3) \right|
$$
5. **Simplify inside parentheses:**
$$
A = \frac{1}{2} \left| (-2)(4) + 4(-4) + 1(0) \right|
$$
6. **Calculate each term:**
$$
A = \frac{1}{2} \left| -8 - 16 + 0 \right| = \frac{1}{2} \left| -24 \right|
$$
7. **Calculate absolute value and multiply:**
$$
A = \frac{1}{2} \times 24 = 12
$$
**Final answer:** The area of triangle $\Delta ABC$ is $12$ square units.