Question

On a map, a hospital is located at (-42,7). A wreck occurs at (13,-14). How far must the Air Care helicopter fly to reach the wreck from the hospital. (Each unit on the map equals one mile.)

Answer 1

Answer:

The helicopter must cover a straight line distance of approximately 61.270 miles from the hospital to the wreck site.

Step-by-step explanation:

Let suppose that Air Care helicopter travels in straight line from the hospital to the wreck site, we can determine that distance ($d$), measured in miles, by using the following Pythagorean identity from Analytical Geometry:

$d = \sqrt{(x_{W}-x_{H})^{2}+(y_{W}-y_{H})^{2}}$

Where:

$x_{W}$, $y_{W}$ - Location of the wreck site, measured in miles.

$x_{H}$, $y_{H}$ - Location of the hospital, measured in miles.

If the location of the wreck site and the hospital are $(13, -14)$ and $(-42, 7)$, respectively. The distance that helicopter must cover is:

$d = \sqrt{[13-(-42)]^{2}+[(-14)-13]^{2}}$

$d \approx 61.270\,mi$

The helicopter must cover a straight line distance of approximately 61.270 miles from the hospital to the wreck site.