# Determine whether the points lie on a straight line

1. Aug 9, 2011

### maff is tuff

1. The problem statement, all variables and given/known data

Determine whether the points lie on a straight line.

a) A(2,4,2), B(3,7,-2), C(1,3,3)

b) D(0,-5,5), E(1,-2,4), F(3,4,2)

2. Relevant equations

3. The attempt at a solution

I tried graphing to see the points A B and C to see if they looked like they were in a straight line and it looked like they were but the answer says they are not in a straight line. I looked on Cramster and they said, "For the points to be in a straight line, the distance between the two sets of points closest to each other must equal the distance between the two points farthest from each other." I am not following this logic. Our solution manual for our class says, "In order for the points to lie on a straight line, the sum of the two shortest distances must be equal to the longest distance." This made a little more sense so I tried to test this by doing it in 2D. I graphed the line y=x and chose 3 points: (0,0), (1,1), and (2,2). It turned out that their method worked for 2D and according to the graph I drew, it looks obvious that the the distance from (0,0) to (1,1) plus the distance from (1,1) to (2,2) equals the distance from (0,0) to (2,2) but I can't seem to visualize/see it graphically in 3D. Any tips? Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 9, 2011

### Staff: Mentor

Do you know how to calculate distance in three dimensions? For problem a, it looks to me like point A is between point C and B. If the points are on the same line, it should be the case that CA + AB = CB.

Last edited: Aug 9, 2011
3. Aug 9, 2011

### maff is tuff

If A is between C and B shouldn't it be: AB + AC = BC? Or am I confused?

4. Aug 9, 2011

### I like Serena

Yep, you're right.
Can you calculate the 3 distances?

5. Aug 9, 2011

### Ray Vickson

You need not compute any distances. All you need to show is that the vector C-A is a scalar multiple of the vector B-A; that is, there is a constant r giving C-A = r*(B-A).

RGV

6. Aug 9, 2011

### Aimless

While the sum of distances calculation that the books suggests is easy and straightforward, I find that I am rather bugged by this approach. I feel like it is more meaningful, in the sense of obtaining a geometric understanding of the problem, to approach the problem differently.

In two dimensions, you can use two points along a line to determine the slope of that line, and then use the slope and one of your two initial points to find the equation the line satisfies (and thus, find all points along the line).

The same approach works in three dimensions, except that you need to reduce the problem into two separate two dimensional problems by projecting the line into the x-y and x-z planes (treat this as calculating an x-y slope and an x-z slope). You then treat the problem exactly as you would in the two dimensional case, except that you will now have two separate equations, and the line is the set of points which satisfies both equations simultaneously.

Edit: Ray Vickson's suggestion above is a more mathematically sophisticated (and computationally easier) method of doing the same thing. If you understand how his approach works (and why it works) then use it. If you don't, then it probably won't help you.

7. Aug 9, 2011

### maff is tuff

Thanks for all th4 replies. Aimless, that makes more sense doing it that way thanks.

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