Question

cameron, arthur, and jamie are playing soccer. their locations are recorded by a motion - tracking system. the distance between gridlines is 5 meters. see examples 3 and 4

1. how far apart are arthur and jamie? round to the nearest tenth of a meter.
2. who is closer to cameron? explain.
3. the soccer ball is located at the point (35, 60). who is closest to the soccer ball?

Explanation:

Step1: Determine Arthur's and Jamie's coordinates

Assume Arthur is at $(20,35)$ and Jamie is at $(45,20)$ (scaled based on 5 - meter gridlines).

Step2: Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

$d=\sqrt{(45 - 20)^2+(20 - 35)^2}=\sqrt{25^2+(- 15)^2}=\sqrt{625 + 225}=\sqrt{850}\approx29.2$ meters.

Step3: For question 21, find distances from Cameron (assume at $(60,35)$) to Arthur and Jamie

Distance from Cameron to Arthur: $d_{CA}=\sqrt{(60 - 20)^2+(35 - 35)^2}=\sqrt{40^2+0^2}=40$ meters.
Distance from Cameron to Jamie: $d_{CJ}=\sqrt{(60 - 45)^2+(35 - 20)^2}=\sqrt{15^2+15^2}=\sqrt{225+225}=\sqrt{450}\approx21.2$ meters. Since $21.2<40$, Jamie is closer to Cameron.

Step4: For question 22, find distances from the ball at $(35,60)$ to Arthur, Jamie and Cameron

Distance from ball to Arthur: $d_{BA}=\sqrt{(35 - 20)^2+(60 - 35)^2}=\sqrt{15^2+25^2}=\sqrt{225 + 625}=\sqrt{850}\approx29.2$ meters.
Distance from ball to Jamie: $d_{BJ}=\sqrt{(35 - 45)^2+(60 - 20)^2}=\sqrt{(-10)^2+40^2}=\sqrt{100 + 1600}=\sqrt{1700}\approx41.2$ meters.
Distance from ball to Cameron: $d_{BC}=\sqrt{(35 - 60)^2+(60 - 35)^2}=\sqrt{(-25)^2+25^2}=\sqrt{625+625}=\sqrt{1250}\approx35.4$ meters. Since $29.2<35.4<41.2$, Arthur is closest to the ball.

Answer:

1. $29.2$ meters
2. Jamie is closer to Cameron because the distance from Cameron to Jamie is approximately $21.2$ meters and the distance from Cameron to Arthur is 40 meters.
3. Arthur is closest to the soccer - ball.