Solve Problem 1.2.10: Points on a Line from Linear Algebra & Its Applications

[Dafe]

Homework Statement

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?

Homework Equations

y = ax + b

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..

 

[Office_Shredder]

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

 

[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?)

 

[Dafe]

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

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}}

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?)

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!

 

[CompuChip]

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

 

[Dafe]

Indeed sir,

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