Find the area of a triangle whose vertices ar | vedclass

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(C) 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

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$24$

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(C) 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 $A(1, -1), B(-4, 6)$ and $C(-3, -5)$.Substituting these values into the formula:Area $= \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$Area $= \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$Area $= \frac{1}{2} |11 + 16 + 21|$Area $= \frac{1}{2} |48| = 24$Thus,the area of the triangle is $24$ square units.

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

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