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In the diagram, \(D\) and \(E\) are the midpoints of \(\overline{AB}\) and \(\overline{BC}\) respectively.

Determine the sum of the \(x\) and \(y\) coordinates of \(F\), the point of intersection of \(\overline{AE}\) and \(\overline{CD}\).

\(\text{Let $\vec{A} =\dbinom06 $} \\ \text{Let $\vec{D} =\dfrac{\vec{A}}{2} = \dbinom03 $} \\ \text{Let $\vec{C} =\dbinom80 $} \\ \text{Let $\vec{E} =\dfrac{\vec{C}}{2} = \dbinom40 $} \\ \text{Let $\vec{F} = \dbinom x y $} \)

\(\begin{array}{|lrcll|} \hline (1) & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \quad \lambda \text{ is a scalar } \\ (2) & \vec{F} &=& \vec{D}+\mu(\vec{C}-\vec{D}) \quad \mu \text{ is a scalar } \\ \hline &\vec{F} = \vec{E}+\lambda(\vec{A}-\vec{E}) &=& \vec{D}+\mu(\vec{C}-\vec{D}) \\ & \mathbf{ \vec{E}+\lambda(\vec{A}-\vec{E}) } &=& \mathbf{\vec{D}+\mu(\vec{C}-\vec{D})} \\ & \lambda(\vec{A}-\vec{E})-\mu(\vec{C}-\vec{D}) &=& \vec{D}-\vec{E} \\\\ &&& \boxed{ \vec{A}-\vec{E} =\dbinom06-\dbinom40=\dbinom{-4}{6} \\ \vec{C}-\vec{D} =\dbinom80-\dbinom03=\dbinom{8}{-3} \\ \vec{D}-\vec{E} =\dbinom03-\dbinom40=\dbinom{-4}{3} }\\\\ & \lambda\dbinom{-4}{6}-\mu\dbinom{8}{-3} &=& \dbinom{-4}{3} \quad | \quad \times \dbinom{3}{8} \\\\ & \lambda\dbinom{-4}{6}\dbinom{3}{8}-\mu\underbrace{\dbinom{8}{-3}\dbinom{3}{8}}_{=0} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda\dbinom{-4}{6}\dbinom{3}{8} &=& \dbinom{-4}{3}\dbinom{3}{8} \\ & \lambda(-12+48) &=& -12+24 \\ & \lambda(36) &=& 12 \quad | \quad : 12 \\ & 3\lambda &=& 1 \\ &\mathbf{ \lambda } &=& \mathbf{ \dfrac{1}{3} } \\ \hline & \vec{F} &=& \vec{E}+\lambda(\vec{A}-\vec{E}) \\ & \vec{F} &=& \dbinom40+\dfrac{1}{3}\dbinom{-4}{6} \\ & \vec{F} &=& \begin{pmatrix} 4-\dfrac{4}{3} \\ \dfrac{6}{3} \end{pmatrix} \\ & \mathbf{\vec{F}} &=& \mathbf{ \begin{pmatrix} \dfrac{8}{3} \\ 2 \end{pmatrix} } \\\\ & x+y &=& \dfrac{8}{3} + 2 \\ &\mathbf{ x+y } &=& \mathbf{\dfrac{14}{3}} \\ \hline \end{array}\)