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, 4), and c(4, −2). 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

Answer 1

To compute for distance from a to b:

Distance A = 4 miles, base of triangle formed between a and b

Distance B = 3 miles, height of triangle formed between a and b

Distance AB = (A^2 + B^2)^0.5 hypotenuse of triangle formed between a and b, or square root of (A^2 + B^2)

Distance AB = ((4)^2 + (3)^2)^0.5

Distance AB = (16 + 9)^0.5

Distance AB = (25)^0.5

Distance AB = 5 miles

To compute for distance from b to where bank employee got stuck

BC = distance from b to c

BC = 4 miles

Total distance driven by bank employee

= AB + 0.5(BC)

= 5 + 0.5(4)

= 5 + 2

= 7 miles, total distance driven by the bank employee

Answer 2

The minimum total distance the employee may have driven before getting stuck in traffic is 9.0 miles.

Explanation:

Given information: a(−3, 1), b(2, 4), and c(4, −2).

On a town map, each unit of the coordinate plane represents 1 mile.

Distance formula:

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

Distance between bank a and bank b is

`ab=√((2-(-3))^2+(4-1)^2)`

`ab=√((5))^2+(3)^2)`

`ab=√(25+9)`

`ab=√(34)`

Distance between bank b and bank c is

`bc=√((4-2)^2+(-2-4)^2)`

`bc=√((2)^2+(-6)^2)`

`bc=√(4+36)`

`bc=√(40)`

`bc=2√(10)`

A bank employee drives from branch a to branch b and then drives halfway to branch c before getting stuck in traffic. It means total distance employee may have driven before getting stuck in traffic is

`Distance=ab+(1)/(2)bc`

`Distance=√(34)+(1)/(2)(2√(10))`

`Distance=√(34)+√(10)`

`Distance=8.99323`

`Distance\approx 9.0`

Therefore the minimum total distance the employee may have driven before getting stuck in traffic is 9.0 miles.