# Points on a line

1. Jul 23, 2009

### Dafe

1. The problem statement, all variables and given/known data
Problem 1.2.10 from Linear Algebra and Its Applications by Gilbert Strang:

Under what condition on $$y_{1}, y_{2}, y_{3}$$ do the points $$(0,y_{1}), (1,y_{2}), (2,y_{3})$$ lie on a straight line?

2. Relevant equations

$$y = ax + b$$

3. The attempt at a solution

If $$y_{1}=0$$,
then
$$y_{2}=y_{1}+1$$
and
$$y_{3}=y_{1}+2$$

This is true, but I guess it's possible to create a lot of straight lines if starting out at the origin.
My attempt at a solution even agrees with the solutions given by Strang.
Am I not reading the problem correctly?
I feel kind of silly for asking this, as it's really basic..

2. Jul 23, 2009

### Office_Shredder

Staff Emeritus
You provide a possible solution, but there are a lot more. What they're really striving for:
Any two points on the plane will give you a unique line. So given y1 and y2, a line will pass through your first two points, and only one such line will. What are the conditions for the third line to lie on the line also? Hint: Think about the slope of the line

3. Jul 23, 2009

### CompuChip

So apparently for the special case y1 = 0 you chose, your solution agrees with the general solution from the book. Now try it for general y1 (i.e. what happens if you shift y1 by a constant c?)

4. Jul 23, 2009

### Dafe

Slope of the first line:

$$a = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

Slope of second line:

$$a = \frac{y_{3}-y_{1}}{x_{3}-x_{1}}$$

The slopes of the two lines must be equal:

$$\frac{y_{2}-y_{1}}{y_{3}-y{1}} = \frac{x_{2}-x_{1}}{x_{3}-x_{1}}$$

I think that;

$$y_{2}-y_{1} = x_{2}-x_{1}$$
and
$$y_{3}-y_{1} = x_{3}-x_{1}$$

are valid for all $$y_{1}$$

Thanks for helping me out guys!

5. Jul 23, 2009

### CompuChip

Yes, now plug in the numbers for xi: you should get the same equations as in your earlier post.

6. Jul 23, 2009

### Dafe

Indeed sir,

$$y_{2}-y_{1} = 1 - 0 = 1$$
$$y_{3}-y_{1} = 2 - 0 = 2$$