If the vertices of Delta ABC are A(-4, -1),B( | vedclass

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

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$17$

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(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(-4, -1)$,$B(1, 2)$,and $C(4, -3)$.Here,$x_1 = -4, y_1 = -1, x_2 = 1, y_2 = 2, x_3 = 4, y_3 = -3$.Substituting these values into the formula:Area $= \frac{1}{2} |-4(2 - (-3)) + 1(-3 - (-1)) + 4(-1 - 2)|$Area $= \frac{1}{2} |-4(5) + 1(-2) + 4(-3)|$Area $= \frac{1}{2} |-20 - 2 - 12|$Area $= \frac{1}{2} |-34|$Area $= \frac{34}{2} = 17$ square units.

If the vertices of $\Delta ABC$ are $A(-4, -1)$,$B(1, 2)$,and $C(4, -3)$,then the area of the triangle is:

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