To find the area of $\triangle ATU$, I need to know the lengths of its base and height. The base of $\triangle ATU$ is the horizontal distance between $T$ and $U$, which is the same as the difference in their $x$-coordinates. To find the $x$-coordinates of $T$ and $U$, I need to use the equations of the lines that they lie on. The equation of the horizontal line that passes through $T$ and $U$ is $y=t$. The equation of the line that passes through $A$ and $B$ is $y=-4x+8$, since it has a slope of $-4$ and a $y$-intercept of $8$. The equation of the line that passes through $A$ and $C$ is $y=-x+8$, since it has a slope of $-1$ and a $y$-intercept of $8$. To find the $x$-coordinate of $T$, I need to solve the system of equations $y=-4x+8$ and $y=t$. Substituting $y=t$ into the first equation, I get $t=-4x+8$, which implies that $x=\frac{8-t}{4}$. To find the $x$-coordinate of $U$, I need to solve the system of equations $y=-x+8$ and $y=t$. Substituting $y=t$ into the second equation, I get $t=-x+8$, which implies that $x=8-t$. Therefore, the base of $\triangle ATU$ is $|8-t-\frac{8-t}{4}|$, which simplifies to $\frac{3}{4}|8-t|$. The height of $\triangle ATU$ is the vertical distance between $A$ and the line $y=t$, which is $|8-t|$. The area of $\triangle ATU$ is $\frac{1}{2}$ times the base times the height, which is $\frac{1}{2} \cdot \frac{3}{4}|8-t| \cdot |8-t|$. This is equal to 13.5, so I have the equation $\frac{3}{8}(8-t)^2=13.5$. Expanding and simplifying, I get $3(8-t)^2=108$, or $(8-t)^2=36$. Taking square roots, I get $8-t=\pm 6$. Adding $t$ to both sides, I get $t=8 \pm 6$. Since $t$ has to be between 0 and 8, the only valid solution is $t=8-6=2$.


Triangle $ABC$ has vertices $A(0, 8)$, $B(2, 0)$, $C(8, 0)$.  A horizontal line with equation $y=t$ intersects line segment $ \overline{AB} $ at $T$ and line segment $ \overline{AC} $ at $U$, forming $\triangle ATU$ with area 13.5.  Compute $t$.