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(3, 3), and C(3, −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 if necessary.

Answer 1

Answer:

The minimum total distance the employee may have driven before getting stuck in traffic is 8.3 miles

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of the distance between to points $(x_{1},y_{1})$

  and $(x_{2},y_{2})$ i s $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

- So at first we will find the distance between the three points

∵ A = (-3 , 1) , B = (3 , 3) , C = (3 , -1)

- By using the rule of distance

∴ $AB=\sqrt{(3--3)^{2}+(3-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325$

∴ $BC=\sqrt{(3-3)^{2}+(-1-3)^{2}}=\sqrt{0+16}=4$

∴ $AC=\sqrt{(3--3)^{2}+(-1-1)^{2}}=\sqrt{36+4}=\sqrt{40}=6.325$

∵ Each unit coordinate plane represents 1 mile

∵ AB = 6.3 units

∴ The distance between branches A and B = 6.325 miles

∵ BC = 4 units

∴ The distance between branches B and C = 4 miles

∵ AC = 6.3 units

∴ The distance between branches A and C = 6.325 miles

- A bank employee drives from Branch A to Branch B and then

 drives halfway to Branch C before getting stuck in traffic

∵ The distance from branch A to branch B is 6.325 miles

∵ The distance from branch B to branch C is 4 miles

∵ The half way from branch B to branch C = 1/2 × 4 = 2 miles

∴ The distance the employee may have driven before getting

   stuck in traffic = 6.325 + 2 = 8.325 miles

∴ The minimum total distance the employee may have driven

   before getting stuck in traffic is 8.3 miles