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(1, 4), and C(5, −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. The minimum total distance the employee may have driven before getting stuck in traffic is

Answer 1

The distance between Branch A and Branch B is given by the Pythagoras theorem

AB² = (Vertical Distance)² + (Horizontal Distance)²
AB² = (4-1)² + (1 - -3)²
AB² = 3² + 4²
AB² = 9 + 16
AB² = 25
AB = 5

BC² = (Vertical distance)² + (Horizontal distance)²
BC² = (-2 - 4)² + (5-1)²
BC² = (-6)² + (4)²
BC² = 36 + 16
BC² = 52
BC = 7.21

Half way of BC = 7.21 ÷ 2 = 3.6 miles

Total distance travelled from A to B and then halfway from B to C is 3.6+5 = 8.6 miles

Answer 2

The total minimum distance traveled before getting stuck in traffic rounded to the nearest tenth of a mile is 8.6 miles

Given the coordinates :

A = (−3, 1)

B = (1, 4)

C = (5, −2)

Recall the Euclidean distance formula :

$d \: = \sqrt{(x_{2} -x_{1})^{2} + (y_{2} -y_{1})^{2} }$

Distance traveled from branch A to branch B :

$x_1 = -3$

$x_2 = 1$

$y_1 = 1$

$y_2 = 4$

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

$d \: = \sqrt{(4)^{2} + (3)^{2}} = \sqrt{16 + 9} = 5$

Distance traveled from branch B to branch C :

• $x_1 = 1$
• $x_2 = 5$
• $y_1 = 4$
• $y_2 = -2$

$d \: = \sqrt{(5 -1)^{2} + ( - 2 - 4)^{2} }$

$d \: = \sqrt{(4)^{2} + ( - 6)^{2}} = \sqrt{16 + 36} = 7.211$

Half the distance between Branch B and C :

$\frac{7.211}{2} = 3.6055 \: miles$

Total minimum distance traveled :

• (Total distance from A to B) + (half distance between B and C)
• $(5 + 3.6055) = 8.6055 = 8.6 \: miles$

Therefore, the total minimum distance traveled is 8.6 miles.

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