SOLUTION: Unit 1 Distance and Midpoint Formulas Geometry Basics ... - Free Printable

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

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🔷 Part 1: Distance Between Two Points

The distance formula 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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#### Problem 1: $(-4, 6)$ and $(3, -7)$

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

$$
\sqrt{218} \approx 14.76 \quad \text{✔ Correct}
$$

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#### Problem 2: $(-6, -5)$ and $(2, 0)$

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

$$
\sqrt{89} \approx 9.43 \quad \text{✔ Correct}
$$

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#### Problem 3: $(-1, 4)$ and $(1, -1)$

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

$$
\sqrt{29} \approx 5.39 \quad \text{✔ Correct}
$$

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#### Problem 4: $(0, -8)$ and $(3, 2)$

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

$$
\sqrt{109} \approx 10.44 \quad \text{✔ Correct}
$$

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#### Problem 5: Graphed points $(-6, -4)$ and $(4, 8)$

From the graph:
- Point A: $(-6, -4)$
- Point B: $(4, 8)$

$$
D = \sqrt{(4 - (-6))^2 + (8 - (-4))^2} = \sqrt{(10)^2 + (12)^2} = \sqrt{100 + 144} = \sqrt{244}
$$

$$
\sqrt{244} \approx 15.62 \quad \text{✘ But student wrote } 12.89
$$

Wait — this is incorrect.

Let’s double-check:

$$
\sqrt{244} = \sqrt{4 \cdot 61} = 2\sqrt{61} \approx 2 \times 7.81 = 15.62
$$

But the student wrote 12.89, which is close to $\sqrt{166}$ or $\sqrt{167}$, but not correct.

👉 Error in Problem 5: The correct distance is approximately 15.62, not 12.89.

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🔷 Part 2: Midpoint Formula

The midpoint formula between $(x_1, y_1)$ 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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#### Problem 6: $A(5, 4)$ and $B(1, -4)$

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

But student wrote $(2, 2)$ ✘

Wait — that’s wrong.

Let’s check:
- $x$: $(5+1)/2 = 3$
- $y$: $(4 + (-4))/2 = 0/2 = 0$

So midpoint is $(3, 0)$, not $(2, 2)$

👉 Error in Problem 6: Should be $(3, 0)$, not $(2, 2)$

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#### Problem 7: $A(3, 8)$ and $B(2, 5)$

$$
M = \left( \frac{3+2}{2}, \frac{8+5}{2} \right) = \left( \frac{5}{2}, \frac{13}{2} \right) = (2.5, 6.5)
$$

Student wrote $\left(\frac{-1}{2}, 6\right)$ ✘

That’s completely wrong.

Wait — maybe a typo?

Let’s recalculate:
- $x = (3 + 2)/2 = 5/2 = 2.5$
- $y = (8 + 5)/2 = 13/2 = 6.5$

So correct answer is $(2.5, 6.5)$ or $\left(\frac{5}{2}, \frac{13}{2}\right)$

Student wrote $\left(-\frac{1}{2}, 6\right)$ → Incorrect

👉 Error in Problem 7

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#### Problem 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 $(0, -6)$ ✘

This is wrong.

- $x$-coordinate: $(5+5)/2 = 5$, not 0
- $y$: $(-7 -5)/2 = -12/2 = -6$

So correct midpoint is $(5, -6)$

👉 Error in Problem 8

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#### Problem 9: $P(2, -6)$ and $Q(4, 5)$

$$
M = \left( \frac{2+4}{2}, \frac{-6 + 5}{2} \right) = \left( \frac{6}{2}, \frac{-1}{2} \right) = (3, -0.5)
$$

Student wrote $(-3, 1)$ ✘

Completely wrong.

Correct: $(3, -0.5)$ or $\left(3, -\frac{1}{2}\right)$

👉 Error in Problem 9

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✔ Summary of Errors:

| Problem | Student Answer | Correct Answer | Status |
|--------|----------------|----------------|--------|
| 5 | 12.89 | ~15.62 | ✘ Incorrect |
| 6 | (2, 2) | (3, 0) | ✘ Incorrect |
| 7 | $(-0.5, 6)$ | $(2.5, 6.5)$ | ✘ Incorrect |
| 8 | (0, -6) | (5, -6) | ✘ Incorrect |
| 9 | (-3, 1) | $(3, -0.5)$ | ✘ Incorrect |

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🛠️ Final Answers (Corrected):

#### Distance Problems:
1. $\sqrt{218} \approx 14.76$ ✔
2. $\sqrt{89} \approx 9.43$ ✔
3. $\sqrt{29} \approx 5.39$ ✔
4. $\sqrt{109} \approx 10.44$ ✔
5. $\sqrt{244} \approx 15.62$ ✘ (student had 12.89)

#### Midpoint Problems:
6. $(3, 0)$ ✘ (student had (2, 2))
7. $\left(\frac{5}{2}, \frac{13}{2}\right) = (2.5, 6.5)$ ✘ (student had $(-0.5, 6)$)
8. $(5, -6)$ ✘ (student had (0, -6))
9. $\left(3, -\frac{1}{2}\right)$ ✘ (student had (-3, 1))

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💡 Tips for Accuracy:

- Always double-check signs when subtracting coordinates.
- Use parentheses: e.g., $y_2 - y_1 = 2 - (-8) = 10$, not $2 - 8 = -6$
- For midpoints: average both coordinates separately.
- Use calculator only after simplifying algebra.

Let me know if you'd like a cleaned-up version of this worksheet with corrections!