The vertices of the triangle ABC are (2, 1),( | vedclass

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(A) The vertices of triangle $ABC$ are $A(2, 1)$,$B(4, 3)$,and $C(2, 5)$.The mid-points $D$,$E$,and $F$ are calculated as follows:$D = \left( \frac{2+

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(A) The vertices of triangle $ABC$ are $A(2, 1)$,$B(4, 3)$,and $C(2, 5)$.The mid-points $D$,$E$,and $F$ are calculated as follows:$D = \left( \frac{2+4}{2}, \frac{1+3}{2} \right) = (3, 2)$$E = \left( \frac{4+2}{2}, \frac{3+5}{2} \right) = (3, 4)$$F = \left( \frac{2+2}{2}, \frac{1+5}{2} \right) = (2, 3)$The area of triangle $DEF$ is given by the determinant formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Area $= \frac{1}{2} |3(4 - 3) + 3(3 - 2) + 2(2 - 4)|$Area $= \frac{1}{2} |3(1) + 3(1) + 2(-2)|$Area $= \frac{1}{2} |3 + 3 - 4| = \frac{1}{2} |2| = 1$ sq. units.Alternatively,the area of the triangle formed by joining the mid-points is $\frac{1}{4}$ of the area of the original triangle $ABC$.

The vertices of the triangle $ABC$ are $(2, 1)$,$(4, 3)$,and $(2, 5)$. If $D$,$E$,and $F$ are the mid-points of the sides,then the area of the triangle $DEF$ is:

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