Question

1. two explorers on an expedition to the arctic circle have radioed their coordinates to base camp. explorer a is at coordinates $(-26,-15)$. explorer b is at coordinates $(13,21)$. the base camp is located at the origin.

a. determine the linear distance between the two explorers.
b. determine the midpoint between the two explorers.
c. determine the distance between the midpoint of the explorers and the base camp.

Explanation:

Step1: Calculate the distance between two explorers (use distance formula)

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Here, \(x_1=-26,y_1=-15,x_2 = 13,y_2=21\).
\(d=\sqrt{(13-(-26))^2+(21-(-15))^2}=\sqrt{(13 + 26)^2+(21+15)^2}=\sqrt{39^2+36^2}\)
\(=\sqrt{1521+1296}=\sqrt{2817}\approx53.08\)

Step2: Calculate the mid - point between two explorers (use mid - point formula)

The mid - point formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\)
\(M=(\frac{-26+13}{2},\frac{-15 + 21}{2})=(-6.5,3)\)

Step3: Calculate the distance between the mid - point and the base camp (use distance formula)

The base camp is at \((0,0)\), and the mid - point is \((-6.5,3)\)
Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(x_1 = 0,y_1=0,x_2=-6.5,y_2 = 3\)
\(d=\sqrt{(-6.5-0)^2+(3 - 0)^2}=\sqrt{42.25+9}=\sqrt{51.25}\approx7.16\)

Answer:

a. The linear distance between the two explorers is approximately \(53.08\) units.
b. The mid - point between the two explorers is \((-6.5,3)\)
c. The distance between the mid - point of the explorers and the base camp is approximately \(7.16\) units.