Long John Silver, a pirate, has buried his treasure on an island with five trees located at the following points: A (30.0m, -20.0m), B(60.0 m, 80.0m), C (-10.0m, -10.0m), D (40.0m, -30.0m), and E(-70.0m, 60.0m). 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-third the distance between your current location and C. Then move toward D, covering one-fourth the distance between where you are and D. Finally, move toward E, covering one-fifth the distance between you and E, stop and dig. (a) What are the coordinates of the point where the pirate's treasure is buried?...after trying several formulas that I know and looking at the internet for help, I haven't been able to find an answer. If anyone can please help me it will be much appreciated.
Vectors and finding points on a line
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algebra-precalculus
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0@joriki: oh, you're absolutely right. I was busy thinking that this is just some computation, and I gave the wrong one. Very silly of me. Thank you. – 2011-09-11
2 Answers
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Hint: The $x$ coordinate of the point a fraction $p$ from $S$ to $T$ is $x_S +p(x_T-x_S)$, and similarly with the $y$ coordinate. Just work through the calculations
Here is a diagram to help (I think)
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0@Ronnie: it really is no more complicated than "starting at $A$ and going half of the way to $B$ from $A$". – 2011-09-11
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The locations are:
P1 = A + (B - A) / 2 P2 = P1 + (C - P1) / 3 P3 = P2 + (D - P2) / 4 P4 = P3 + (E - P3) / 5 == treasure location
Each equation are in fact 2 equations for the x and y coordinates respectively.
B - A
is the vector from A to B. When you are in A and move along the vector B - A
you arrive at A + (B - A) = B
. When you move only half of the distance you arrive at P1 = A + (B - A)/2
.
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0Yes, finally it's good to remember that B-A is the vector from A to B. But it is also easy to understand. In order to get from A to B you need to find (x, y) such that (Ax, Ay) + (x, y) = (Bx, By). You solve and get (x, y) = (Bx-Ax, By-Ay), and this is B-A. – 2011-09-11