# Vector parametric equation of a line

1. Oct 5, 2008

### roam

Find a vector parametric equation of the line in $$R^{2}$$ with equation 2x-3y = 4

Attempt at a solution

I haven't seen this type of question before so I don't know where to start. I suppose that the equation 2x-3y = 4 is a vector equation of that line and is in the form x = x0 + tv. I just don't know how to split this vector equation up and try to write the parametric vector equation.

Can someone please explain to me what to do. Many Thanks.

2. Oct 5, 2008

### danago

2x - 3y = 4 is the cartesian equation, not the vector equation.

If a line has the vector equation r = x0 + tv, then it passes through the point x0 and is parallel to the vector v.

Can you find a point that the line 2x - 3y = 4 passes through? Can you find a vector to which it is parallel?

3. Oct 5, 2008

### tiny-tim

Hi roam!

Yes, that is the right form.

x0 can be any point on the line.

So just choose some point on the line (at random), and chug away.

4. Oct 5, 2008

### HallsofIvy

Staff Emeritus
One of your difficulties is that there are many different parametric equations for the same line.

Yes, you want x= at+ b, y= ct+ d. But you can choose "t" anyway you want. You might, for example, choose t to be x itself: that is x= t. Now y= ?

5. Oct 5, 2008

### roam

Hi Tim!

I think I should even write the equation 2x-3y=4 in the form:

$$y = \frac{2x -4}{3}$$ for some t

And I choose some random point on the line, (2,0)

Now for x = x0 + bt, y = y0+ bt

If x = t, like Hall said, y = $$\frac{2t -4}{3}$$

is this the vector parametric equation of the line? Do I need to specify a "t" as well?

6. Oct 5, 2008

### roam

Is it (t, $$\frac{2t -4}{3}$$) ?

7. Oct 5, 2008

### tiny-tim

vector parametric equation

Hi roam!
So far, so good.

(though you could have got that straight from 2x-3y=4, couldn't you? )

At this point I'm going to disagree with HallsofIvy, and say that I think the question is asking for a pure vector equation, rather than a coordinate-based one.

Coordinate-based would be of the form x = f(t), y = g(t).

Pure vector would be of the form r(t) = a + bt, where a and b are both constant vectors.

So the answer would be r(t) = (2,0) + (?,?)t.

8. Oct 7, 2008

### roam

Re: vector parametric equation

Hi Tim,

How do we determine the direction vector? Thanks!

9. Oct 7, 2008

### tiny-tim

oh come on!

just look at the original equation: 2x-3y = 4 …

if x goes up by 1, then y goes up by … ?

10. Oct 7, 2008

### HallsofIvy

Staff Emeritus
Very little difference between "parametric equations", x= f(t), y= g(t), as I gave, and "vector parametric equation", $\vec{r}= f(t)\vec{i}+ g(t)\vec{j}$.

11. Oct 7, 2008

### roam

Would it be 3/2?

12. Oct 7, 2008

### BoundByAxioms

Hmm, try again.

You yourself said that y=$$\frac{2x-4}{3}$$. What is the slope of that line? How would you express that slope in the form of a direction vector?

13. Oct 7, 2008

### roam

m = 2/3 ?

14. Oct 7, 2008

### roam

$$y = \frac{2x-4}{3}$$
if x = 1 then y = -10/3

Is that it? can we plug it in the equation (2,0) + t(*,*)?

15. Oct 7, 2008

### tiny-tim

Yes!
uh? wots happening?

(x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2).

16. Oct 7, 2008

### BoundByAxioms

You're golden.

17. Oct 7, 2008

### BoundByAxioms

If x=1, then y=-2/3. Just for clarification. Where did you get the -10/3 from?