Question

you are helping your city plan emergency routes in a new neighborhood. the neighborhood is on a coordinate grid where each square represents 100 meters. emergency vehicles can travel on streets (grid lines) or cut diagonally through parks when responding to emergencies. question 6 design your own emergency facility location anywhere on the grid. explain why you chose this spot using at least two distance calculations.

Explanation:

Let's assume we choose the point \((5,5)\) as the emergency facility location.

1. Distance to the fire - truck location \((2,1)\):

• Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), where \((x_1,y_1)=(2,1)\) and \((x_2,y_2)=(5,5)\).
• \(d=\sqrt{(5 - 2)^2+(5 - 1)^2}=\sqrt{9 + 16}=\sqrt{25}=5\) units. Since each unit is \(100\) meters, the distance is \(5\times100 = 500\) meters.

1. Distance to the hospital - like location \((8,2)\):

• Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), where \((x_1,y_1)=(5,5)\) and \((x_2,y_2)=(8,2)\).
• \(d=\sqrt{(8 - 5)^2+(2 - 5)^2}=\sqrt{9+9}=\sqrt{18}\approx4.24\) units. In meters, it is approximately \(4.24\times100 = 424\) meters.
• The point \((5,5)\) is relatively central. It is not too far from the fire - truck (which may need to cover a large area horizontally and vertically in case of fires) and also has a reasonable distance to the hospital - like location (which may receive patients from various directions).

Answer:

We can choose the point \((5,5)\) as the emergency facility location. The distance to the fire - truck at \((2,1)\) is \(500\) meters (using the distance formula \(d = 5\times100\) where \(d=\sqrt{(5 - 2)^2+(5 - 1)^2}=5\) units) and the distance to the hospital - like location at \((8,2)\) is approximately \(424\) meters (using the distance formula \(d\approx4.24\times100\) where \(d=\sqrt{(8 - 5)^2+(2 - 5)^2}\approx4.24\) units). This location is relatively central, allowing for efficient response times to different types of emergencies.