To find the area of triangle ABC given the points A(-7, 3), B(5, 8), and C(0, -2), we can use the formula for the area of a triangle based on its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Where \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are the coordinates of the vertices A, B, and C, respectively.
Substituting the coordinates of points A, B, and C:
- A(-7, 3) → \( x_1 = -7, y_1 = 3 \)
- B(5, 8) → \( x_2 = 5, y_2 = 8 \)
- C(0, -2) → \( x_3 = 0, y_3 = -2 \)
Using the formula, we compute the area:
\[ \text{Area} = \frac{1}{2} \left| -7(8 - (-2)) + 5((-2) - 3) + 0(3 - 8) \right| \]
Calculating each term:
- \( -7(8 + 2) = -7 \times 10 = -70 \)
- \( 5(-2 - 3) = 5 \times -5 = -25 \)
- \( 0(3 - 8) = 0 \)
Now, plug these values into the area formula:
\[ \text{Area} = \frac{1}{2} \left| -70 - 25 + 0 \right| = \frac{1}{2} \left| -95 \right| = \frac{1}{2} \times 95 = 47.5 \]
Thus, the area of triangle ABC is 47.5 square units.