Let L be the line y = 5-9x in R^2. Find vectors a and b so that v = a +tb is a parametric equation for L.

Question

Stuck on this linear algebra question.

Answer 1

Let v = (v₁, v₂), let a = (a₁, a₂), and let b = (b₁, b₂).

 

Then we can write two equations from v = a + tb. (v₁, v₂) = (a₁, a₂) + t(b₁, b₂), so:

 

v₁ = a₁ + tb₁

v₂ = a₂ + tb₂

 

We'd like (v₁, v₂) to be a point on L for t any real number. When t = 0, v₁ = a₁ and v₂ = a₂. That means that (a₁, a₂) should just be some point on L. We could pick any point, but let's make it the y-intercept, (0, 5). So, a = (0, 5).

 

 

Now, let's think about b. b represents the slope of your line in some way, since it tells you what to do every time you increase t. That is, every time you increase t, you're adding some multiple of b to your origin vector a. This is like the slope of line L, since every time you increase x, you're adding some multiple of the slope to your y-intercept.

 

L has a slope of -9, or -9/1, so it moves down 9 for every 1 it moves right. That means we'd probably like to add 1 to our x-coordinate v₁ and -9 to our y-coordinate v₂ every time we increase t by 1. So, let b = (1, -9).

 

 

To see that at work, plug in a = (0, 5) and b = (1, -9) and see what happens when t = 1:

 

v = (0, 5) + (1)(1, -9)

v = (0, 5) + (1, -9)

v = (1, -4)

 

Is that a point on L?

 

y = 5 - 9x

-4 = 5 - 9(1)

-4 = 5 - 9

-4 = -4

 

That checks out!