imc

Let $A$ be at $(6,5)$, B be at $(8,-3)$, and $C$ be at $(9,1)$. Reflecting over the line $x=8$, we see that $A^{\prime }=D=(10,5)$, $B^{\prime }=B$ (as the x-coordinate of B is 8), and $C^{\prime }=E=(7,1)$. Line $AB$ can be represented as $y=-4x+29$, so we see that $E$ is on line $AB$. We see that if we connect $A$ to $D$, we get a line of length $4$ (between $(6,5)$ and $(10,5)$). The area of $△ABD$ is equal to $\frac{bh}{2}=\frac{4(8)}{2}=16$. Now, let the point of intersection between $AC$ and $DE$ be $F$. If we can just find the area of $△ADF$ and subtract it from $16$, we are done. We realize that because the diagram is symmetric over $x=8$, the intersection of lines $AC$ and $DE$ should intersect at an x-coordinate of $8$. We know that the slope of $DE$ is $\frac{5-1}{10-7}=\frac{4}{3}$. Thus, we can represent the line going through $E$ and $D$ as $y-1=\frac{4}{3}(x-7)$. Plugging in $x=8$, we find that the y-coordinate of F is $\frac{7}{3}$. Thus, the height of $△ADF$ is $5-\frac{7}{3}=\frac{8}{3}$. Using the formula for the area of a triangle, the area of $△ADF$ is $\frac{16}{3}$. To get our final answer, we must subtract this from $16$. $[ABD]-[ADF]=16-\frac{16}{3}=\boxed{\frac{32}{3}}$