Question

After obtaining a warrant, the police attached GPS tracking device to the car of a murder suspect. Suppose the car was located at the origin of a Cartesian coordinate system when the device was attached. Shortly afterwards, the suspect's car was tracked going 9 mi due east, mi due north, and 1 mi due west before coming to a permanent stop. (a) What are the coordinates of the suspect's car at its final destination? (x, y) - (b) What was the distance traveled by the suspect? X mi (c) What is the distance as the crow flies between the original position and the final position of the suspect's car?

Answer 1

Answer:

Part a) The coordinates of the suspect's car at its final destination are (4,4)

Part b) The distance traveled by the suspect was 10 miles

Part c) $d=5.66\ mi$

Step-by-step explanation:

The correct question in the attached figure

Part a) What are the coordinates of the suspect's car at its final destination?

we know that

The original position is the origin of a Cartesian coordinate system (0,0)

1) The suspect's car was tracked going 5 mi due east

That means----> 5 mi at right

the new coordinates are (5,0)

2) The suspect's car was tracked going 4 mi due north

That means----> 4 mi up

the new coordinates are (5,4)

3) The suspect's car was tracked going 1 mi due west

That means----> 1 mi at left

the new coordinates are (4,4)

therefore

The coordinates of the suspect's car at its final destination are (4,4)

Part b) What was the distance traveled by the suspect?

The distance traveled is  

5 mi due east + 4 mi due north + 1 mi due west

$5+4+1=10\ mi$

therefore

The distance traveled by the suspect was 10 miles

Part c)  What is the distance as the crow flies between the original position and the final position of the suspect's car?

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Remember that

The original position was (0,0) and the final position was (4,4)

substitute in the formula

$d=\sqrt{(4-0)^{2}+(4-0)^{2}}$

$d=\sqrt{(4)^{2}+(4)^{2}}$

$d=\sqrt{32}\ mi$

$d=5.66\ mi$