To find the area of triangle ABC with vertices A(-7, 3), B(5, 8), and C(0, -2), we will use the formula for the area of a triangle given its vertex coordinates:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Assigning the coordinates:
- \( A(x_1, y_1) = (-7, 3) \)
- \( B(x_2, y_2) = (5, 8) \)
- \( C(x_3, y_3) = (0, -2) \)
Now substituting these coordinates into the formula:
\[ \text{Area} = \frac{1}{2} \left| -7(8 - (-2)) + 5((-2) - 3) + 0(3 - 8) \right| \]
Calculating each term:
-
\( 8 - (-2) = 8 + 2 = 10 \)
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\( -7 \cdot 10 = -70 \)
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\( -2 - 3 = -5 \)
-
\( 5 \cdot -5 = -25 \)
So, now substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} \left| -70 - 25 + 0 \right| = \frac{1}{2} \left| -95 \right| = \frac{1}{2} \cdot 95 = 47.5 \]
Thus, the area of triangle ABC is \( 47.5 \) square units.
The correct answer is 47.5.