## Linear Algebra and Its Applications, exercise 1.2.13

Exercise 1.2.13. Given the equation x + 4y = 7 for a line in the x-y plane, find the equation for a line that is parallel to the first line and passes through the point (0, 0). The first line passes through the point (1, 3) (3, 1); find another line that also passes through that point.

Answer: Solving for y in the first equation, we have $x + 4y = 7 \Rightarrow 4y = -x + 7 \Rightarrow y = - \frac{1}{4}x + \frac{7}{4}$

This line has slope 1/4 and intersects the y axis at the point (0, 7/4). The parallel line (also of slope 1/4) passing through the origin has the equation $y = - \frac{1}{4}x$.

If we assume a slope of 1 then the corresponding line has an equation of the form y = x + a for some a. If we further assume that the line goes through the point (1, 3) (3, 1) then a = 2 a = -2. So the line with equation y = x + 2 y = x – 2 satisfies the condition for the second part of the exercise.

UPDATE: Corrected the point of intersection. (I had transcribed the exercise incorrectly.) Thanks to freaky4you 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 1.2.13

1. freaky4you says:

Meeting Point is (3,1) not (1,3) so the answer will be y=x-2

• hecker says:

Fixed. Thanks for finding and reporting the error!