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:
The minimum total distance the employee may have driven before getting stuck in traffic is 9.0 miles.
Step-by-step 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:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Distance between bank a and bank b is
[tex]ab=\sqrt{(2-(-3))^2+(4-1)^2}[/tex]
[tex]ab=\sqrt{(5))^2+(3)^2}[/tex]
[tex]ab=\sqrt{25+9}[/tex]
[tex]ab=\sqrt{34}[/tex]
Distance between bank b and bank c is
[tex]bc=\sqrt{(4-2)^2+(-2-4)^2}[/tex]
[tex]bc=\sqrt{(2)^2+(-6)^2}[/tex]
[tex]bc=\sqrt{4+36}[/tex]
[tex]bc=\sqrt{40}[/tex]
[tex]bc=2\sqrt{10}[/tex]
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
[tex]Distance=ab+\frac{1}{2}bc[/tex]
[tex]Distance=\sqrt{34}+\frac{1}{2}(2\sqrt{10})[/tex]
[tex]Distance=\sqrt{34}+\sqrt{10}[/tex]
[tex]Distance=8.99323[/tex]
[tex]Distance\approx 9.0[/tex]
Therefore the minimum total distance the employee may have driven before getting stuck in traffic is 9.0 miles.
