Question

Find a Parametrization for the curve: The line through the points
3, 3 and
5, 1.
Would the answer be: y = 8t+3 & x=4t-3

Answer 1

Do you mean "Would AN answer be...?" There is more than one parameterization.

To get y = 0, your formulation requires t = 0.
t = 0 gives x = -3

You seem to have missed the point!

Try y = mt + 3 and x = nt + 3
Now, for t = 0, we hit (3,3)

How do we choose n and m so as to find (5,1)?

Answer 2

u have to points of the line u=(3,3) and v=(5,1) so the direction of the line is
u-v=(3,3)-(5,1)=(-2,2)

so u can write the line as
\[ \large \{(3,3)+t(-2,2):t\in\mathbb{R}\} \]
this means that
\[ \large x=3-2t\qquad y=3+2t \]

Answer 3

Yea I ment is my answer a possible answer

Answer 4

u can check if your answer is right by solving
\[ \large 3=8t+3\qquad 3=4t-3 \]
u should get the same t from both equations.

Answer 5

So it is right?