Coordinate Geometry Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each problem step-by-step using the Distance Formula and Midpoint Formula as provided in your worksheet.

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🔹 Distance Formula

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

We'll use this to find the distance between two points.

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🔹 Midpoint Formula

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

We’ll use this to find the midpoint between two points.

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## ✔ Part 1: Distance Problems

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1. (-3, 5) and (-7, -8)

$$
d = \sqrt{(-7 - (-3))^2 + (-8 - 5)^2} = \sqrt{(-4)^2 + (-13)^2} = \sqrt{16 + 169} = \sqrt{185}
$$

✔ Answer: $\boxed{\sqrt{185}}$ or approximately $13.60$

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2. (-5, 8) and (-4, 1)

$$
d = \sqrt{(-4 - (-5))^2 + (1 - 8)^2} = \sqrt{(1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}
$$

✔ Answer: $\boxed{5\sqrt{2}}$ or approximately $7.07$

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3. (5, 4) and (3, 8)

$$
d = \sqrt{(3 - 5)^2 + (8 - 4)^2} = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
$$

✔ Answer: $\boxed{2\sqrt{5}}$ or approximately $4.47$

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4. (0, -6) and (-7, 1)

$$
d = \sqrt{(-7 - 0)^2 + (1 - (-6))^2} = \sqrt{(-7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}
$$

✔ Answer: $\boxed{7\sqrt{2}}$ or approximately $9.90$

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5. (2, -3) and (8, -5)

$$
d = \sqrt{(8 - 2)^2 + (-5 - (-3))^2} = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
$$

✔ Answer: $\boxed{2\sqrt{10}}$ or approximately $6.32$

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6. (-4, -2) and (7, -3)

$$
d = \sqrt{(7 - (-4))^2 + (-3 - (-2))^2} = \sqrt{(11)^2 + (-1)^2} = \sqrt{121 + 1} = \sqrt{122}
$$

✔ Answer: $\boxed{\sqrt{122}}$ or approximately $11.05$

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## ✔ Part 2: Midpoint Problems

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1. (8, 0) and (-9, 1)

$$
x = \frac{8 + (-9)}{2} = \frac{-1}{2} = -0.5,\quad y = \frac{0 + 1}{2} = \frac{1}{2} = 0.5
$$

✔ Answer: $\boxed{(-0.5, 0.5)}$ or $\boxed{\left(-\frac{1}{2}, \frac{1}{2}\right)}$

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2. (-9, -10) and (4, -9)

$$
x = \frac{-9 + 4}{2} = \frac{-5}{2} = -2.5,\quad y = \frac{-10 + (-9)}{2} = \frac{-19}{2} = -9.5
$$

✔ Answer: $\boxed{(-2.5, -9.5)}$ or $\boxed{\left(-\frac{5}{2}, -\frac{19}{2}\right)}$

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3. (4, -3) and (2, 9)

$$
x = \frac{4 + 2}{2} = \frac{6}{2} = 3,\quad y = \frac{-3 + 9}{2} = \frac{6}{2} = 3
$$

✔ Answer: $\boxed{(3, 3)}$

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4. (8, 7) and (-8, 3)

$$
x = \frac{8 + (-8)}{2} = \frac{0}{2} = 0,\quad y = \frac{7 + 3}{2} = \frac{10}{2} = 5
$$

✔ Answer: $\boxed{(0, 5)}$

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5. (7, -8) and (-2, 8)

$$
x = \frac{7 + (-2)}{2} = \frac{5}{2} = 2.5,\quad y = \frac{-8 + 8}{2} = \frac{0}{2} = 0
$$

✔ Answer: $\boxed{(2.5, 0)}$ or $\boxed{\left(\frac{5}{2}, 0\right)}$

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

$$
x = \frac{-4 + 3}{2} = \frac{-1}{2} = -0.5,\quad y = \frac{7 + (-1)}{2} = \frac{6}{2} = 3
$$

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

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## 📌 Final Answers Summary

✔ Distance Problems:

| Problem | Answer |
|--------|--------|
| 1 | $\sqrt{185}$ |
| 2 | $5\sqrt{2}$ |
| 3 | $2\sqrt{5}$ |
| 4 | $7\sqrt{2}$ |
| 5 | $2\sqrt{10}$ |
| 6 | $\sqrt{122}$ |

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✔ Midpoint Problems:

| Problem | Answer |
|--------|--------|
| 1 | $\left(-\frac{1}{2}, \frac{1}{2}\right)$ |
| 2 | $\left(-\frac{5}{2}, -\frac{19}{2}\right)$ |
| 3 | $(3, 3)$ |
| 4 | $(0, 5)$ |
| 5 | $\left(\frac{5}{2}, 0\right)$ |
| 6 | $\left(-\frac{1}{2}, 3\right)$ |

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Let me know if you'd like these answers formatted for a printed worksheet!