Question

Mr. Brady is using a coordinate plane to design a treasure hunt for his students. The hunt begins at the flagpole. The first clue is hidden 5 units north of the flagpole. The second clue is located 6 units east of the flagpole. Clue 2 says that Clue 3 is located 5 units south of the flagpole. N -4 -2 o N 4 6 X -2 -4 Plot the locations of the flagpole and the 3 clues on the coordinate grid and show the path students will follow with straight lines. Each unit represents 50 feet. What is the shortest combined distance along the path from the flagpole to Clue 1 to Clue 2 to Clue 3? Round to the nearest foot if necessary. ​

Answer 1

The figure is attached and the total distance is 1031 feet.

Explanation:

The graph is indicated in the attached figure.

For calculation of distance consider following

Point Flagpole is (0,0)

Point Clue1 is (0,5)

Point Clue2 is (6,0)

Point Clue3 is (0,-5)

So the distance is calculated as follows

`d_(T)=[d_(FP\ to\ Clue1)+d_(Clue1\ to\ Clue2)+d_(Clue2\ to\ Clue3)]*distance\ per\ unit\\d_(T)=[√((Clue1_x-FP_x)^2+(Clue1_y-FP_y)^2)+√((Clue2_x-Clue1_x)^2+(Clue2_y-Clue1_y)^2)+√((Clue3_x-Clue2_x)^2+(Clue3_y-Clue2_y)^2)]**distance\ per\ unit\\`Substituting the values

`d_(T)=[√((0-0)^2+(5-0)^2)+√((6-0)^2+(0-5)^2)+√((0-6)^2+(-5-0)^2)]*50 \text{ feet}\\d_(T)=[√((0)^2+(5)^2)+√((6)^2+(-5)^2)+√((-6)^2+(-5)^2)]*50 \text{ feet}\\d_(T)=[√(0+25)+√(36+25)+√(36+25)]*50 \text{ feet}\\d_(T)=[√(25)+√(61)+√(61)]*50 \text{ feet}\\d_(T)=[5+7.81+7.81]*50 \text{ feet}\\d_(T)=[20.62]*50 \text{ feet}\\d_(T)=1031 \text{ feet}\\`

So the total distance travelled is 1031 feet.