Completed homework on distance and midpoint formulas with handwritten solutions.

Let's go through the homework step by step and verify each answer, explaining the Distance Formula and Midpoint Formula as needed.

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🔹 Distance Formula

The distance $ D $ between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

$$
D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

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🔹 Midpoint Formula

The midpoint $ M $ of a segment with endpoints $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

$$
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$

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## ✔ Part 1: Find the Distance Between Each Pair of Points

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1. (-4, 6) and (3, -7)

$$
D = \sqrt{(3 - (-4))^2 + (-7 - 6)^2} = \sqrt{(3 + 4)^2 + (-13)^2} = \sqrt{7^2 + 169} = \sqrt{49 + 169} = \sqrt{218}
$$

$$
\sqrt{218} \approx 14.76
$$

✔ Correct! The student wrote 14.76, which matches.

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2. (-6, -5) and (2, 0)

$$
D = \sqrt{(2 - (-6))^2 + (0 - (-5))^2} = \sqrt{(8)^2 + (5)^2} = \sqrt{64 + 25} = \sqrt{89}
$$

$$
\sqrt{89} \approx 9.43
$$

✔ Correct! Student wrote 9.43 — good.

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3. (-1, 4) and (1, -1)

$$
D = \sqrt{(1 - (-1))^2 + (-1 - 4)^2} = \sqrt{(2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}
$$

$$
\sqrt{29} \approx 5.39
$$

✔ Correct! Student wrote 5.39 — correct.

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4. (0, -8) and (3, 2)

$$
D = \sqrt{(3 - 0)^2 + (2 - (-8))^2} = \sqrt{3^2 + 10^2} = \sqrt{9 + 100} = \sqrt{109}
$$

$$
\sqrt{109} \approx 10.44
$$

✔ Correct! Student wrote 10.44 — accurate.

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5. Graphed points: (-6, -1) and (4, 6)

From the graph:
- Point A: $(-6, -1)$
- Point B: $(4, 6)$

$$
D = \sqrt{(4 - (-6))^2 + (6 - (-1))^2} = \sqrt{(10)^2 + (7)^2} = \sqrt{100 + 49} = \sqrt{149}
$$

$$
\sqrt{149} \approx 12.21
$$

✔ Student wrote 12.21 — correct!

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## ✔ Part 2: Find the Midpoint

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6. P(5, 4) and Q(1, -4)

$$
M = \left( \frac{5 + 1}{2}, \frac{4 + (-4)}{2} \right) = \left( \frac{6}{2}, \frac{0}{2} \right) = (3, 0)
$$

✘ Incorrect! Student wrote (2, 2) — this is wrong.

✔ Correct Answer: (3, 0)

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7. M(1, 9) and N(2, 3)

$$
M = \left( \frac{1 + 2}{2}, \frac{9 + 3}{2} \right) = \left( \frac{3}{2}, \frac{12}{2} \right) = (1.5, 6)
$$

✔ Student wrote $ \left(\frac{3}{2}, 6\right) $ — correct!

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8. P(5, -7) and Q(5, -5)

$$
M = \left( \frac{5 + 5}{2}, \frac{-7 + (-5)}{2} \right) = \left( \frac{10}{2}, \frac{-12}{2} \right) = (5, -6)
$$

✘ Student wrote (0, -6) — incorrect.

✔ Correct Answer: (5, -6)

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9. P(2, -6) and Q(4, 8)

$$
M = \left( \frac{2 + 4}{2}, \frac{-6 + 8}{2} \right) = \left( \frac{6}{2}, \frac{2}{2} \right) = (3, 1)
$$

✘ Student wrote (-3, 1) — incorrect.

✔ Correct Answer: (3, 1)

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## ✔ Final Review & Corrections

| Problem | Student's Answer | Correct Answer | Status |
|--------|------------------|----------------|--------|
| 1 | 14.76 | √218 ≈ 14.76 | ✔ Correct |
| 2 | 9.43 | √89 ≈ 9.43 | ✔ Correct |
| 3 | 5.39 | √29 ≈ 5.39 | ✔ Correct |
| 4 | 10.44 | √109 ≈ 10.44 | ✔ Correct |
| 5 | 12.21 | √149 ≈ 12.21 | ✔ Correct |
| 6 | (2, 2) | (3, 0) | ✘ Incorrect |
| 7 | (3/2, 6) | (1.5, 6) | ✔ Correct |
| 8 | (0, -6) | (5, -6) | ✘ Incorrect |
| 9 | (-3, 1) | (3, 1) | ✘ Incorrect |

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## 🛠️ Explanation of Mistakes

✘ Problem 6:

- Points: (5, 4) and (1, -4)
- x-coordinate: $ \frac{5+1}{2} = 3 $, not 2.
- y-coordinate: $ \frac{4 + (-4)}{2} = 0 $
- So midpoint is (3, 0), not (2, 2)

✘ Problem 8:

- Both x-coordinates are 5 → average is $ \frac{5+5}{2} = 5 $
- y-values: $ \frac{-7 + (-5)}{2} = -6 $
- So midpoint is (5, -6), not (0, -6)

✘ Problem 9:

- $ \frac{2+4}{2} = 3 $, not -3
- $ \frac{-6+8}{2} = 1 $
- So midpoint is (3, 1), not (-3, 1)

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## ✔ Final Answers (Corrected)

Distance Problems:

1. 14.76
2. 9.43
3. 5.39
4. 10.44
5. 12.21

Midpoint Problems:

6. (3, 0)
7. $ \left(\frac{3}{2}, 6\right) $ or (1.5, 6)
8. (5, -6)
9. (3, 1)

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