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

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