The vertices of Delta ABC are A(2, 4),B(5, 0) | 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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$6$

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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 $A(2, 4)$,$B(5, 0)$,and $C(5, 4)$.Substituting the values into the formula:Area $= \frac{1}{2} |2(0 - 4) + 5(4 - 4) + 5(4 - 0)|$Area $= \frac{1}{2} |2(-4) + 5(0) + 5(4)|$Area $= \frac{1}{2} |-8 + 0 + 20|$Area $= \frac{1}{2} |12| = 6$ square units.Alternatively,since $AC$ is a horizontal line (both have $y=4$) and $BC$ is a vertical line (both have $x=5$),the triangle is a right-angled triangle at $C$.Base $AC = |5 - 2| = 3$ units.Height $BC = |4 - 0| = 4$ units.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 4 = 6$ square units.

The vertices of $\Delta ABC$ are $A(2, 4)$,$B(5, 0)$,and $C(5, 4)$. Then the area of $\Delta ABC = \ldots$

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