The area of a triangle having vertices A(3, 0 | 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

$4.5$

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 $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$.Substituting these values into the formula:Area $= \frac{1}{2} |3(3 - 3) + 0(3 - 0) + 3(0 - 3)|$Area $= \frac{1}{2} |3(0) + 0(3) + 3(-3)|$Area $= \frac{1}{2} |0 + 0 - 9|$Area $= \frac{1}{2} |-9| = \frac{9}{2} = 4.5$ square units.

The area of a triangle having vertices $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$ is:

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