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Based on the information provided in the image, we need to determine the location of a property such that the distances to three specific locations are equal:

1. The wife's workplace located at the intersection of Calle 6 and Calle E.
2. His office located at the intersection of Calle 2 and Calle E.
3. The children's school located at the intersection of Calle 4 and Calle A.

We can observe that each block is a square and moving from one street to another (either from Calle 1 to Calle 2 or from Calle A to Calle B) is considered a unit distance.

To simplify, let’s break down the distances:
- **From Calle 6 to Calle 2**: 4 units
- **From Calle 6 to Calle 4**: 2 units
- **From Calle E to Calle A**: 4 units

Given that the property must be equidistant from these three locations, let’s analyze the possible options:

- **Option A (Calle 4 and Calle C)**:
  - To Calle 6 and Calle E: \(|6-4| + |E-C| = 2 + 2 = 4\) units
  - To Calle 2 and Calle E: \(|2-4| + |E-C| = 2 + 2 = 4\) units
  - To Calle 4 and Calle A: \(|4-4| + |A-C| = 0 + 2 = 2\) units
- **Option B (Calle 4 and Calle D)**:
  - To Calle 6 and Calle E: \(|6-4| + |E-D| = 2 + 1 = 3\) units
  - To Calle 2 and Calle E: \(|2-4| + |E-D| = 2 + 1 = 3\) units
  - To Calle 4 and Calle A: \(|4-4| + |A-D| = 0 + 3 = 3\) units
- **Option C (Calle 4 and Calle E)**:
  - To Calle 6 and Calle E: \(|6-4| = 2\) units
  - To Calle 2 and Calle E: \(|2-4| = 2\) units
  - To Calle 4 and Calle A: \(|4-4| + |A-E| = 4\) units
- **Option D (Calle 5 and Calle C)**:
  - To Calle 6 and Calle E: \(|6-5| + |E-C| = 1 + 2 = 3\) units
  - To Calle 2 and Calle E: \(|2-5| + |E-C| = 3 + 2 = 5\) units
  - To Calle 4 and Calle A: \(|4-5| + |A-C| = 1 + 2 = 3\) units

Based on this analysis, **Option B (Calle 4 and Calle D)** ensures that the property is equidistant (3 units) from all three destinations. Therefore, the correct answer to the problem is **B) Calle 4 y Calle D**.

If you need more details or have any other questions, please let me know! Here are some related questions you might consider:

1. How would the solution change if the distances to one of the points were to be prioritized?
2. What would be the impact of different street configurations on the problem solution?
3. How does the grid layout facilitate the calculation of distances?
4. Could there be multiple correct solutions if the distances were calculated differently?
5. How does city planning influence daily commutes based on this problem?

**Tip**: Remember that in a grid system, the Manhattan distance (sum of the absolute differences of their Cartesian coordinates) is used to calculate distances rather than direct linear distance.

Don Gerardo decided to move to a property located at an equal distance from his wife's workplace (Calle 6 and Calle E), his office (Calle 2 and Calle E), and his children's school (Calle 4 and Calle A). The question is to determine at which intersection the property should be located.

Solve a distance problem on a grid to help Don Gerardo find a property equidistant from his wife's workplace, his office, and his children's school. Learn how to calculate distances using the Manhattan Distance formula.

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