Question

eric, cam, and brooke are playing mini golf on a computer. the game stores objects in terms of x- and y-coordinates.
the graph below shows the position of each players ball and the hole marked x at (3, 2).
graph with coordinate plane, erics ball at (-6, 6), brookes at (10, 8), cams at (8, -6), hole x at (3, 2), graph units in meters
what is the distance to the hole for the player who is farthest from the hole?
round the final answer to the nearest tenth of a meter. do not round intermediate calculations.
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Explanation:

Step1: Identify coordinates

First, we need to find the coordinates of each player's ball. From the graph:

• Eric's ball: Let's assume the coordinates. Looking at the x - axis (left - right) and y - axis (up - down), Eric's ball is at \((- 6,6)\) (since it's 6 units left on x - axis and 6 units up on y - axis).
• Brooke's ball: From the graph, it is at \((10,8)\) (10 units right on x - axis and 8 units up on y - axis).
• Cam's ball: From the graph, it is at \((8,-6)\) (8 units right on x - axis and 6 units down on y - axis).

The hole (X) is at \((3,2)\).

Step2: Use distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

Distance for Eric:

\(x_1=-6,y_1 = 6,x_2 = 3,y_2=2\)
\(d_{Eric}=\sqrt{(3-(-6))^2+(2 - 6)^2}=\sqrt{(9)^2+(-4)^2}=\sqrt{81 + 16}=\sqrt{97}\approx9.8\)

Distance for Brooke:

\(x_1 = 10,y_1=8,x_2 = 3,y_2 = 2\)
\(d_{Brooke}=\sqrt{(3 - 10)^2+(2 - 8)^2}=\sqrt{(-7)^2+(-6)^2}=\sqrt{49+36}=\sqrt{85}\approx9.2\)

Distance for Cam:

\(x_1=8,y_1=-6,x_2 = 3,y_2 = 2\)
\(d_{Cam}=\sqrt{(3 - 8)^2+(2-(-6))^2}=\sqrt{(-5)^2+(8)^2}=\sqrt{25 + 64}=\sqrt{89}\approx9.4\)

Step3: Compare distances

Now we compare the distances:

• \(d_{Eric}\approx9.8\)
• \(d_{Brooke}\approx9.2\)
• \(d_{Cam}\approx9.4\)

So the farthest distance is Eric's distance to the hole, which is \(\sqrt{97}\approx9.8\) meters.

Answer:

\(9.8\)