Find the area of the triangle whose vertices | vedclass

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(B) 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(

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) 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 $(x_1, y_1) = (-5, -1)$,$(x_2, y_2) = (3, -5)$,and $(x_3, y_3) = (5, 2)$.Substituting these values into the formula:Area $= \frac{1}{2} |(-5)(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))|$Area $= \frac{1}{2} |(-5)(-7) + 3(3) + 5(4)|$Area $= \frac{1}{2} |35 + 9 + 20|$Area $= \frac{1}{2} |64| = 32$ square units.

Find the area of the triangle whose vertices are (in square units): $(-5,-1), (3,-5), (5,2)$.

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