To find the area of triangle ABC with vertices at A (-5, -4), B (3, -3), and C (3, 2), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
Plugging in the coordinates for points A, B, and C:
- \(x_1 = -5\), \(y_1 = -4\) (point A)
- \(x_2 = 3\), \(y_2 = -3\) (point B)
- \(x_3 = 3\), \(y_3 = 2\) (point C)
We can now substitute these values into the formula:
\[ \text{Area} = \frac{1}{2} \left| -5(-3-2) + 3(2-(-4)) + 3(-4-(-3)) \right| \]
Calculating each term step by step:
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Calculate \(-5(-3-2)\): \[ -5(-5) = 25 \]
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Calculate \(3(2 - (-4))\): \[ 3(2 + 4) = 3 \times 6 = 18 \]
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Calculate \(3(-4 - (-3))\): \[ 3(-4 + 3) = 3(-1) = -3 \]
Now we sum these results:
\[ 25 + 18 - 3 = 40 \]
Now substitute this back into the area formula:
\[ \text{Area} = \frac{1}{2} |40| = \frac{40}{2} = 20 \]
Thus, the area of Leonardo's triangle is 20 square units.