Question

cameron, arthur, and jamie are playing soccer. their locations are recorded by a motion - tracking system. the grid shows distances in meters. who is closer to cameron? explain. click the icon to view the soccer field. arthur is about meters away from cameron, and jamie is about meters away from cameron, so is closer to cameron. (round to the nearest whole number as needed.)

Explanation:

Step1: Determine coordinates

Assume Cameron: $(60,40)$, Arthur: $(20,30)$, Jamie: $(40,20)$

Step2: Use distance formula for Arthur - Cameron

$d_{CA}=\sqrt{(20 - 60)^2+(30 - 40)^2}=\sqrt{1700}\approx41$

Step3: Use distance formula for Jamie - Cameron

$d_{CJ}=\sqrt{(40 - 60)^2+(20 - 40)^2}=\sqrt{800}\approx28$

Step4: Compare distances

Since $28<41$, Jamie is closer.

Answer:

Let's assume Cameron's coordinates are $(x_{C},y_{C})$, Arthur's coordinates are $(x_{A},y_{A})$ and Jamie's coordinates are $(x_{J},y_{J})$. From the grid, if we assume Cameron is at $(60,40)$, Arthur is at $(20,30)$ and Jamie is at $(40,20)$.
The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the distance - formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

• Distance between Cameron and Arthur:
• Substitute $x_1 = 60,y_1 = 40,x_2 = 20,y_2 = 30$ into the distance formula.
• $d_{CA}=\sqrt{(20 - 60)^2+(30 - 40)^2}=\sqrt{(-40)^2+(-10)^2}=\sqrt{1600 + 100}=\sqrt{1700}\approx41$ (rounded to the nearest whole - number).
• Distance between Cameron and Jamie:
• Substitute $x_1 = 60,y_1 = 40,x_2 = 40,y_2 = 20$ into the distance formula.
• $d_{CJ}=\sqrt{(40 - 60)^2+(20 - 40)^2}=\sqrt{(-20)^2+(-20)^2}=\sqrt{400+400}=\sqrt{800}\approx28$ (rounded to the nearest whole - number).

Arthur is about $41$ meters away from Cameron, and Jamie is about $28$ meters away from Cameron, so Jamie is closer to Cameron.