Vector Form: Line Passing Through (2, -1) and Parallel to 2x-3y=1

[Math9999]

Homework Statement

Find the vector form of the equation of the line in ℝ² that passes through P=(2, -1) and is parallel to the line with general equation 2x-3y=1.

Homework Equations

None.

The Attempt at a Solution

vector form: x=p+td
general form: ax+by=c
--------------------------------
a=2, b=-3, so [x, y]=[2, -1]+t[2, -3], am I right?

 

[PeroK]

Homework Statement

Find the vector form of the equation of the line in ℝ² that passes through P=(2, -1) and is parallel to the line with general equation 2x-3y=1.

Homework Equations

None.

The Attempt at a Solution

vector form: x=p+td
general form: ax+by=c
--------------------------------
a=2, b=-3, so [x, y]=[2, -1]+t[2, -3], am I right?

That looks like a guess. Can you draw those two lines and see whether they are parallel or not?

 

[Math9999]

They are but how do I figure out the normal vector?

 

[PeroK]

They are but how do I figure out the normal vector?

They are not parallel.

 

[Math9999]

So I thought that the normal vector should be 2 and -3 based on the general form on the problem but now I don't think that's true. So how do I figure out the normal vector?

 

[Mark44]

a=2, b=-3, so [x, y]=[2, -1]+t[2, -3], am I right?

You can (and should) check this yourself. Clearly, the line from your equation goes through the point (2, -1). Do you see why? But does it have the right slope?
PeroK is saying that the line you calculated and the given line aren't parallel, which should suggest to you that your slope isn't right.

As a check on your work, find the equation of the line you're after, using the point-slope form of the equation of the line. That line should agree with the one you've calculated using a vector form.

So how do I figure out the normal vector?

Why do you need the normal vector?

 

[Math9999]

So 2x-3y=1, 3y=2x-1, y=2/3x-1/3 and m=2/3, which is the slope?

 

[Mark44]

So 2x-3y=1, 3y=2x-1, y=2/3x-1/3 and m=2/3, which is the slope?

Yes, that's the right slope. Now you should be able to get the equation in vector form.

 

[Math9999]

But how? I still don't get it. If the right slope is m=2/3, then how is the answer [x, y]=[2, -1]+t[3, 2]?

 

[Mark44]

But how? I still don't get it. If the right slope is m=2/3, then how is the answer [x, y]=[2, -1]+t[3, 2]?

What's the slope of the vector <3, 2>? Try drawing it in the plane.

 

[Math9999]

m=rise/run=2/3?

 

[Mark44]

m=rise/run=2/3?

Yes
From post #1

Find the vector form of the equation of the line in ℝ2 that passes through P=(2, -1) and is parallel to the line with general equation 2x-3y=1.

Post #9

If the right slope is m=2/3, then how is the answer [x, y]=[2, -1]+t[3, 2]?

Is there some value of t for which <x, y> = <2, -1>? IOW, does the line pass through (2, -1)?
Is the line represented by your vector equation parallel to the line whose equation is 2x - 3y = 1?
.