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Worksheet for calculating distances between pairs of points on a coordinate plane using the Pythagorean theorem.

Graph showing coordinate points on a grid with labels A through Z, used for calculating distances using the Pythagorean theorem.

Graph showing coordinate points on a grid with labels A through Z, used for calculating distances using the Pythagorean theorem.

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Show Answer Key & Explanations Step-by-step solution for: Calculating the Distance Between Two Points Using Pythagorean ...
Let's solve the problem step by step using the Pythagorean distance formula:

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

We will calculate the distance between each pair of points on the coordinate grid to the nearest hundredth.

---

Step 1: Identify coordinates of each point from the graph



From the image, we read off the coordinates of each labeled point:

- A: (8, 3)
- B: (0, -6)
- C: (-5, 8)
- D: (-1, 4)
- E: (-2, 0)
- F: (-9, -7)
- G: (0, -3)
- H: (0, -9)
- J: (2, -4)
- K: (0, 1)
- M: (-4, -5)
- N: (-3, -8)
- P: (2, 9)
- Q: (-5, 7)
- R: (4, 2)
- S: (-5, 0)
- T: (8, -9)
- V: (6, -3)
- W: (-3, -7)
- Z: (1, 0)

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Now compute each distance:

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1. $ d(A, B) $



A(8, 3), B(0, -6)

$$
d = \sqrt{(0 - 8)^2 + (-6 - 3)^2} = \sqrt{(-8)^2 + (-9)^2} = \sqrt{64 + 81} = \sqrt{145} \approx 12.04
$$

12.04

---

2. $ d(E, F) $



E(-2, 0), F(-9, -7)

$$
d = \sqrt{(-9 + 2)^2 + (-7 - 0)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.90
$$

9.90

---

3. $ d(J, K) $



J(2, -4), K(0, 1)

$$
d = \sqrt{(0 - 2)^2 + (1 + 4)^2} = \sqrt{(-2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39
$$

5.39

---

4. $ d(P, Q) $



P(2, 9), Q(-5, 7)

$$
d = \sqrt{(-5 - 2)^2 + (7 - 9)^2} = \sqrt{(-7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \approx 7.28
$$

7.28

---

5. $ d(T, V) $



T(8, -9), V(6, -3)

$$
d = \sqrt{(6 - 8)^2 + (-3 + 9)^2} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32
$$

6.32

---

6. $ d(C, D) $



C(-5, 8), D(-1, 4)

$$
d = \sqrt{(-1 + 5)^2 + (4 - 8)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66
$$

5.66

---

7. $ d(G, H) $



G(0, -3), H(0, -9)

$$
d = \sqrt{(0 - 0)^2 + (-9 + 3)^2} = \sqrt{0 + (-6)^2} = \sqrt{36} = 6.00
$$

6.00

---

8. $ d(M, N) $



M(-4, -5), N(-3, -8)

$$
d = \sqrt{(-3 + 4)^2 + (-8 + 5)^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16
$$

3.16

---

9. $ d(R, S) $



R(4, 2), S(-5, 0)

$$
d = \sqrt{(-5 - 4)^2 + (0 - 2)^2} = \sqrt{(-9)^2 + (-2)^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.22
$$

9.22

---

10. $ d(W, Z) $



W(-3, -7), Z(1, 0)

$$
d = \sqrt{(1 + 3)^2 + (0 + 7)^2} = \sqrt{(4)^2 + (7)^2} = \sqrt{16 + 49} = \sqrt{65} \approx 8.06
$$

8.06

---

Final Answers:



$$
\begin{align*}
d(A,B) &= 12.04 \\
d(E,F) &= 9.90 \\
d(J,K) &= 5.39 \\
d(P,Q) &= 7.28 \\
d(T,V) &= 6.32 \\
d(C,D) &= 5.66 \\
d(G,H) &= 6.00 \\
d(M,N) &= 3.16 \\
d(R,S) &= 9.22 \\
d(W,Z) &= 8.06 \\
\end{align*}
$$

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All distances are rounded to the nearest hundredth as requested.

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