A(6,1), B(8,2) and C(9,4) are three vertices | vedclass

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(A) Given that $A(6,1), B(8,2)$ and $C(9,4)$ are three vertices of a parallelogram $ABCD$.Let the fourth vertex $D$ be $(x, y)$.Since the diagonals of

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$\frac{3}{4}$

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(A) Given that $A(6,1), B(8,2)$ and $C(9,4)$ are three vertices of a parallelogram $ABCD$.Let the fourth vertex $D$ be $(x, y)$.Since the diagonals of a parallelogram bisect each other,the midpoint of $BD$ is the same as the midpoint of $AC$.$	ext{Midpoint of } AC = \left(\frac{6+9}{2}, \frac{1+4}{2}ight) = \left(\frac{15}{2}, \frac{5}{2}ight)$.$	ext{Midpoint of } BD = \left(\frac{8+x}{2}, \frac{2+y}{2}ight)$.Equating the coordinates: $\frac{8+x}{2} = \frac{15}{2} \Rightarrow x = 7$ and $\frac{2+y}{2} = \frac{5}{2} \Rightarrow y = 3$.So,the fourth vertex is $D(7,3)$.$E$ is the midpoint of $DC$,so $E = \left(\frac{7+9}{2}, \frac{3+4}{2}ight) = \left(8, \frac{7}{2}ight)$.The area of $	riangle ADE$ with vertices $A(6,1), D(7,3)$ and $E(8, 3.5)$ is given by:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$	ext{Area} = \frac{1}{2} |6(3 - 3.5) + 7(3.5 - 1) + 8(1 - 3)|$$	ext{Area} = \frac{1}{2} |6(-0.5) + 7(2.5) + 8(-2)|$$	ext{Area} = \frac{1}{2} |-3 + 17.5 - 16|$$	ext{Area} = \frac{1}{2} |-1.5| = 0.75 = \frac{3}{4} 	ext{ sq. units.}$

$A(6,1), B(8,2)$ and $C(9,4)$ are three vertices of a parallelogram $ABCD$. If $E$ is the midpoint of $DC$,find the area of $\triangle ADE$ (in sq. units).

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