Solve Long John Silver's Pirate Treasure Puzzle: Find Coordinates

[rcwha]

Long John Silver, a pirate, has buried his treasure on an island with five trees located at the following points: A (30.0 m, -20.0 m), B (60.0 m, 80.0 m), C (-10.0 m, -10.0 m), D (40.0 m, -30.0 m), and E (-70.0 m, 60.0 m). All of the points are measured relative to some origin. Long John's map instructs you to start at A and move toward B, but cover only one-half the distance between A and B. then move toward C, covering one-half the distance between your current location and C. Then move toward D, covering one-fifth the distance between where you are and D. Finally, move toward E, covering one-fifth the distance between you and E, stop, and dig.

What are the coordinates of the point where the pirate's treasure is buried?

i added all the x's and y's (following the rules of above) and got x 49 and y = 21

i got the problem wrong..so what am i doing wrong?


the question also says that you can rearrange the order of the trees - for instance, B (30 m, -20 m), A (60 m, 80 m), E (-10 m, -10 m), C (40 m, -30 m), and D (-70 m, 60 m) - and repeat the calculation to show that the answer does not depend on the order...i did the latter and still got the problem wrong...

please help

 

[Doc Al]

Why don't you spell out your intermediate steps along the way so we can see where you might have made an error.

 

[rcwha]

k...so i have
Ax = 30
Ay= -20

Bx = 60
By = 80 half the distance from A to B would be adding (60/2) to Ax and (80/2) to Ay...so now I am at (60,20)

Cx = -10
Cy = -10 half the distance from (60,20) to C would be adding (-10/2) to 60 and (-10/2) to 20...so now I am at (55,15)

dx = 40 (1/5) the distance from (55,15) to D gives me (63,9)
dy = -30

Ex = -70
Ey = 60 (1/5) the distance from (63,9) to E gives me (49,21) as my final point

 

[teken894]

The keyword here is "half the distance from A to B"

You are dividing the coordinate of point B by two, then adding to A..which is wrong.

You first need to find the distance, or the VECTOR AB

broken down into components:
Bx - Ax = ABx
By - Ay = ABy

which is simply:

60-30 = 30 = ABx
80-(-20)=100 = ABy
vecotr AB = <30,100>

Now, divide both the vector by two, or multiply by (1/2)

vectorAB/2 will result in the components <15,50>

now take that vector and add that to the coordinate of point A

( (30+15), (-20+50) ) = (45,30) is the point halfway from point A to B, or "half the distance from A to B"



Of course you could simply use the midpoint forumula

pt = ((x1+x2)/2, (y1+y2)/2))..and that will be the same result...

 

[rcwha]

k i followed ur steps and got (3.6,13.6)...can somebody please confirm my answer...also, when finding the (1/5th) distance between two points could you use (x1+x2)/5, y1+y2/5...i tried this but i got conflicting numbers compared to doing it the other way...

 

[Doc Al]

...also, when finding the (1/5th) distance between two points could you use (x1+x2)/5, y1+y2/5

No. Adding 1/5 of the distance (just like adding half the distance) means:
x1 + (x2 - x1)/5; y1 + (y2 - y1)/5.

Of course, adding half the distance is a special case, since:
x1 + (x2 - x1)/2 = (x1 + x2)/2, but that won't work in general.

 

[rcwha]

like i said earlier, i got (3.6,13.6) as my final answer from using tekken's method...did i get it right?

 

[Doc Al]

i got (3.6,13.6) as my final answer

Looks good to me.