To solve the problem:

Step 1: Calculate the midpoints of $AB$ and $AC$

The midpoint formula is: $\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Midpoint $D$ of $AB$:

$D = \left( \frac{-1 + 3}{2}, \frac{6 + 2}{2} \right) = \left( 1, 4 \right)$

Midpoint $E$ of $AC$:

$E = \left( \frac{-1 + (-5)}{2}, \frac{6 + (-6)}{2} \right) = \left( -3, 0 \right)$

Step 2: Calculate the midpoint of $DE$

Using the midpoint formula again: $\text{Midpoint of } DE = \left( \frac{x_D + x_E}{2}, \frac{y_D + y_E}{2} \right)$ $\text{Midpoint of } DE = \left( \frac{1 + (-3)}{2}, \frac{4 + 0}{2} \right) = \left( -1, 2 \right)$

Step 3: Calculate the length of $DE$

The distance formula is: $\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the coordinates of $D(1, 4)$ and $E(-3, 0)$: $\text{Length of } DE = \sqrt{(-3 - 1)^2 + (0 - 4)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Final Answer:

1. The midpoint of $DE$ is $(-1, 2)$.
2. The length of $DE$ is $4\sqrt{2}$.

Would you like further clarification or a visualization? Here are 5 related questions to deepen your understanding:

1. What would the length of $DE$ be if $A, B,$ or $C$ shifted coordinates?
2. How does the midpoint formula differ when applied in three dimensions?
3. What is the significance of the line $DE$ in terms of the triangle's geometry?
4. Can you verify the calculations by plotting the points on a graph?
5. How is the distance formula derived from the Pythagorean theorem?

Tip: Always label your axes and points when solving graphically to avoid confusion.