# Thread: Cartesan equation to parametric and vector form

1. ## Cartesan equation to parametric and vector form

Please check and please explain if it is incorrect, even if it is correct. have a very vague understanding of it. Thank you.

Convert the following Cartesan equation to parametric and vector form
3x-5y=9
y=-9/5+3/5x
y=-6/5 + 3/5(x-1)
y=-6/5+3/5t
t=x-1

x=t+1
y=-6/5+3/5t
How to write it in a vector form?
Thank you

2. Here is one way.
The the points $\displaystyle (3,0)~\&~(-2,-3)$ are on the line. (You can pick any two.)
Therefore, the vector they determine, $\displaystyle <5,3>$, directs the line.
So the parametric form is $\displaystyle \begin{gathered} x = 5t + 3 \hfill \\ y = 3t \hfill \\ \end{gathered}$

3. Plato,

thank you and I have questions. Why vector is <5,3>, and not <-5, -3>
If we select two points (3,0) and (-2,-3), then v=<-2-3, -3-0>=<-5,-3>
the vector form is -5i-3j
I know it is wrong because your answer is different, but why?

4. Originally Posted by oceanmd
Why vector is <5,3>, and not <-5, -3>
Don't you understand that both $\displaystyle <5,3>~\&~<-5,-3>$ are the same direction of that line?
They are negatives of each other, therefore parallel.
The vectors $\displaystyle <15,9>~\&~<-10,-6>$ are also direction vectors of that line.
Any vector parallel to $\displaystyle <5,3>$ is a direction vector of the line.

5. ## parametric form question

Plato,
got you. Thanks.
As for the parametric form, is it a rule that if a vector is let say <7,5>, then parametric form is x=7t+5 and y=5t ?

Thank you