# Vector parametric equation of a line

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.

danago
Gold Member
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?

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

the form x = x0 + tv.

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. HallsofIvy
Homework Helper
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= ?

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?

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

tiny-tim
Homework Helper
vector parametric equation

Hi 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)

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

How do we determine the direction vector? Thanks!

tiny-tim
Homework Helper
Hi Tim,

How do we determine the direction vector? Thanks!

oh come on! just look at the original equation: 2x-3y = 4 …

if x goes up by 1, then y goes up by … ? HallsofIvy
Homework Helper
Very little difference between "parametric equations", x= f(t), y= g(t), as I gave, and "vector parametric equation", $\vec{r}= f(t)\vec{i}+ g(t)\vec{j}$.

oh come on! just look at the original equation: 2x-3y = 4 …

if x goes up by 1, then y goes up by … ? Would it be 3/2?

Would it be 3/2?

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?

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?

m = 2/3 ?

if x goes up by 1, then y goes up by … ? $$y = \frac{2x-4}{3}$$
if x = 1 then y = -10/3

Is that it? can we plug it in the equation (2,0) + t(*,*)? tiny-tim
Homework Helper
m = 2/3 ?

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

Is that it? can we plug it in the equation (2,0) + t(*,*)? uh? wots happening? (x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2). Yes! uh? wots happening? (x,y) = (2,0) + t(1,2/3), or perhaps more neatly (x,y) = (2,0) + t(3,2). You're golden.

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

Is that it? can we plug it in the equation (2,0) + t(*,*)? If x=1, then y=-2/3. Just for clarification. Where did you get the -10/3 from?