Student's homework on distance and midpoint formulas in geometry.

Problem Analysis:

The task involves solving two types of problems:
1. Finding the distance between pairs of points using the distance formula.
2. Finding the midpoint of line segments defined by pairs of points using the midpoint formula.

Let's solve each part step by step.

---

Part 1: Distance Formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### Problem 1: Points \((-4, 6)\) and \((3, -7)\)
\[
D = \sqrt{(3 - (-4))^2 + (-7 - 6)^2}
\]
\[
D = \sqrt{(3 + 4)^2 + (-7 - 6)^2}
\]
\[
D = \sqrt{7^2 + (-13)^2}
\]
\[
D = \sqrt{49 + 169}
\]
\[
D = \sqrt{218}
\]
\[
D \approx 14.76
\]

#### Problem 2: Points \((-6, -5)\) and \((2, 0)\)
\[
D = \sqrt{(2 - (-6))^2 + (0 - (-5))^2}
\]
\[
D = \sqrt{(2 + 6)^2 + (0 + 5)^2}
\]
\[
D = \sqrt{8^2 + 5^2}
\]
\[
D = \sqrt{64 + 25}
\]
\[
D = \sqrt{89}
\]
\[
D \approx 9.43
\]

#### Problem 3: Points \((-1, 4)\) and \((1, -1)\)
\[
D = \sqrt{(1 - (-1))^2 + (-1 - 4)^2}
\]
\[
D = \sqrt{(1 + 1)^2 + (-1 - 4)^2}
\]
\[
D = \sqrt{2^2 + (-5)^2}
\]
\[
D = \sqrt{4 + 25}
\]
\[
D = \sqrt{29}
\]
\[
D \approx 5.39
\]

#### Problem 4: Points \((0, -8)\) and \((3, 2)\)
\[
D = \sqrt{(3 - 0)^2 + (2 - (-8))^2}
\]
\[
D = \sqrt{3^2 + (2 + 8)^2}
\]
\[
D = \sqrt{3^2 + 10^2}
\]
\[
D = \sqrt{9 + 100}
\]
\[
D = \sqrt{109}
\]
\[
D \approx 10.44
\]

---

Part 2: Midpoint Formula

The midpoint formula for two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

#### Problem 5: Points \((3, 5)\) and \((1, 1)\)
\[
\text{Midpoint} = \left( \frac{3 + 1}{2}, \frac{5 + 1}{2} \right)
\]
\[
\text{Midpoint} = \left( \frac{4}{2}, \frac{6}{2} \right)
\]
\[
\text{Midpoint} = (2, 3)
\]

#### Problem 6: Points \((6, 4)\) and \((4, 2)\)
\[
\text{Midpoint} = \left( \frac{6 + 4}{2}, \frac{4 + 2}{2} \right)
\]
\[
\text{Midpoint} = \left( \frac{10}{2}, \frac{6}{2} \right)
\]
\[
\text{Midpoint} = (5, 3)
\]

#### Problem 7: Points \((7, 7)\) and \((3, -1)\)
\[
\text{Midpoint} = \left( \frac{7 + 3}{2}, \frac{7 + (-1)}{2} \right)
\]
\[
\text{Midpoint} = \left( \frac{10}{2}, \frac{6}{2} \right)
\]
\[
\text{Midpoint} = (5, 3)
\]

#### Problem 8: Points \((2, 5)\) and \((4, 1)\)
\[
\text{Midpoint} = \left( \frac{2 + 4}{2}, \frac{5 + 1}{2} \right)
\]
\[
\text{Midpoint} = \left( \frac{6}{2}, \frac{6}{2} \right)
\]
\[
\text{Midpoint} = (3, 3)
\]

---

Final Answers:

1. Distance between \((-4, 6)\) and \((3, -7)\): \(\boxed{14.76}\)
2. Distance between \((-6, -5)\) and \((2, 0)\): \(\boxed{9.43}\)
3. Distance between \((-1, 4)\) and \((1, -1)\): \(\boxed{5.39}\)
4. Distance between \((0, -8)\) and \((3, 2)\): \(\boxed{10.44}\)

5. Midpoint of \((3, 5)\) and \((1, 1)\): \(\boxed{(2, 3)}\)
6. Midpoint of \((6, 4)\) and \((4, 2)\): \(\boxed{(5, 3)}\)
7. Midpoint of \((7, 7)\) and \((3, -1)\): \(\boxed{(5, 3)}\)
8. Midpoint of \((2, 5)\) and \((4, 1)\): \(\boxed{(3, 3)}\)