Find the area (in square units) of a triangle whose vertices are (1, -1), (-4, 6), and (-3, -5).

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Accepted Answer

The formula for the area of a triangle given its vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, and $$(x_3, y_3)$$ is:  
$$
	ext{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ight|
$$

Substituting the given points:  
- $$A(1, -1)$$, $$B(-4, 6)$$, $$C(-3, -5)$$:  
$$
	ext{Area} = \frac{1}{2} \left[ 1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6) ight]
$$

Simplify each term:  
- $$1(6 + 5) = 11$$,  
- $$(-4)(-5 + 1) = -16$$,  
- $$(-3)(-1 - 6) = 21$$.

Add these values:  
$$
	ext{Area} = \frac{1}{2} \left[ 11 - 16 + 21 ight] = \frac{1}{2} 	imes 16 = 24.
$$

Thus, the area of the triangle is **24 square units**, and the correct answer is **(b)**.

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