Worksheet with five word problems related to distance and midpoint calculations using coordinate geometry.

Since I can't view or access images directly, I’ll solve the problems based on your text description. Here are the solutions to each of the word problems involving distance and the coordinate plane.

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Problem 1:

Julie's house is at (−2, −5), and Jimmy's house is at (6, −5). How long is the direct path from Julie’s house to Jimmy’s house?

#### Solution:
We use the distance formula:
$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

Given:
- Julie: $(-2, -5)$
- Jimmy: $(6, -5)$

Plug in:
$$
\text{Distance} = \sqrt{(6 - (-2))^2 + (-5 - (-5))^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8
$$

✔ Answer: The direct path is 8 units long.

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Problem 2:

The Riley and Brown families decided to go to a concert together. The Riley's live 6 miles west and 3 miles south of the concert. The Browns live 2 miles east and 4 miles south. How far apart are they?

#### Step 1: Assign coordinates
Let the concert location be the origin: $(0, 0)$

- Riley's house:
6 miles west → $x = -6$
3 miles south → $y = -3$
So, Riley: $(-6, -3)$

- Brown's house:
2 miles east → $x = 2$
4 miles south → $y = -4$
So, Brown: $(2, -4)$

#### Step 2: Use distance formula
$$
\text{Distance} = \sqrt{(2 - (-6))^2 + (-4 - (-3))^2} = \sqrt{(8)^2 + (-1)^2} = \sqrt{64 + 1} = \sqrt{65}
$$

✔ Answer: They are $\boxed{\sqrt{65}}$ miles apart, or approximately 8.06 miles.

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Problem 3:

A command center learns that an enemy patrol is located 7 miles east and 13 miles north of their position. If the command center's helicopter is located 1 mile east and 2 miles south of the center, what is the shortest distance the helicopter can travel to get to the enemy patrol?

#### Step 1: Set up coordinates
Let the command center be at $(0, 0)$

- Enemy patrol:
7 miles east → $x = 7$
13 miles north → $y = 13$
So, enemy: $(7, 13)$

- Helicopter:
1 mile east → $x = 1$
2 miles south → $y = -2$
So, helicopter: $(1, -2)$

#### Step 2: Distance between helicopter and enemy
$$
\text{Distance} = \sqrt{(7 - 1)^2 + (13 - (-2))^2} = \sqrt{(6)^2 + (15)^2} = \sqrt{36 + 225} = \sqrt{261}
$$

Simplify:
$$
\sqrt{261} = \sqrt{9 \times 29} = 3\sqrt{29}
$$

✔ Answer: The shortest distance is $\boxed{3\sqrt{29}}$ miles, or approximately 16.16 miles.

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Problem 4:

Walk City is located at (−1, −3) and Rushville is located at (4, 9). How long is the route if a train travels along a straight line from Walk City to Rushville? (One map unit equals one mile.)

Use the distance formula:

- Walk City: $(-1, -3)$
- Rushville: $(4, 9)$

$$
\text{Distance} = \sqrt{(4 - (-1))^2 + (9 - (-3))^2} = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
$$

✔ Answer: The route is 13 miles long.

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Problem 5:

Coach Avarado placed Kaiten at (1, 3) and Jeremy at (−6, 5). What is the distance, in yards, of the path from Kaiten to Jeremy?

Assuming the coordinate units are in yards, we use the distance formula:

- Kaiten: $(1, 3)$
- Jeremy: $(-6, 5)$

$$
\text{Distance} = \sqrt{(-6 - 1)^2 + (5 - 3)^2} = \sqrt{(-7)^2 + (2)^2} = \sqrt{49 + 4} = \sqrt{53}
$$

✔ Answer: The distance is $\boxed{\sqrt{53}}$ yards, or approximately 7.28 yards.

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✔ Final Answers Summary:

1. 8 units
2. $\boxed{\sqrt{65}}$ miles ≈ 8.06 miles
3. $\boxed{3\sqrt{29}}$ miles ≈ 16.16 miles
4. 13 miles
5. $\boxed{\sqrt{53}}$ yards ≈ 7.28 yards

Let me know if you'd like these visualized or explained further!