## 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 .

This entry was posted in linear algebra. Bookmark the permalink.

### 2 Responses to Linear Algebra and Its Applications, exercise 1.2.10

1. coruja38 says:

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

• hecker says:

Fixed. Thanks!