Linear Algebra and Its Applications, exercise 1.2.10

Exercise 1.2.10. Assume we have the points $(0, y_1)$ , $(1, y_2)$ , and $(2, y_3)$ . What values must $y_1$ , $y_2$ , and $y_3$ have in order for these points to fall on a straight line?

Answer: If the points fall on the same line then we must have the slope $\Delta y / \Delta x$ be the same between the first two points as it is between the second two points. We therefore have

$(y_2 - y_1) / (1 - 0) = (y_3 - y_2) / (2 - 1) \Rightarrow y_2 - y_1 = y_3 - y_2$

UPDATE: Corrected the formula for the slope; thanks to coruja38 for the fix!

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, exercise 1.2.10

coruja38 says:

January 17, 2013 at 3:34 am

Small mistake. The slope \Delta x / \Delta y instead of \Delta y / \Delta x.

- hecker says:
  
  January 18, 2013 at 2:30 am
  
  Fixed. Thanks!