Question

On a town map, each unit of the coordinate plane represents 1 mile. Three branches of a bank are located at A(-3,1), B(2,3), and C(4,-1). A bank employee drives from branch A to branch B and then drives halfway to branch C before getting stuck in traffic. What is the minimum total distance the employee may have driven before getting stuck in traffic? Round to the nearest tenth of a mile.

Answer 1

we have

A(-3,1), B(2,3), and C(4,-1)  

we know that

the distance between two points is equal to the formula

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

so

Step 1

Find the distance AB

$d=\sqrt{(3-1)^{2}+(2+3)^{2}}$

$d=\sqrt{(2)^{2}+(5)^{2}}$

$dAB=\sqrt{29}\ units$

convert units to miles

$dAB=\sqrt{29}\ miles$

Step 2

find the distance BC

$dBC=\sqrt{(-1-3)^{2}+(4-2)^{2}}$

$dBC=\sqrt{20}\\dBC=2\sqrt{5}\ units$

convert unit to miles

$dBC=2\sqrt{5}\ miles$

Step 3

Find the minimum total distance the employee may have driven before getting stuck in traffic

Sum the distance AB and the half distance BC

$=\sqrt{29}\ miles+(0.5)*2\sqrt{5}\ miles\\ =5.4+2.2\\ =7.6\ miles$

therefore

the answer is

$7.6\ miles$