The line $y = \frac{5}{3} x - \frac{17}{3}$ is to be parameterized using vectors.  Which of the following options are valid parameterizations?

(A) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} -3 \\ -5 \end{pmatrix}$

(B) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 17 \\ 5 \end{pmatrix} + t \begin{pmatrix} 6 \\ 10 \end{pmatrix}$

(C) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -7/3 \end{pmatrix} + t \begin{pmatrix} 3/5 \\ 1 \end{pmatrix}$

(D) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 14/5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 3/5 \end{pmatrix}$

(E) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ -17/3 \end{pmatrix} + t \begin{pmatrix} 15 \\ -25 \end{pmatrix}$

Enter the letters of the correct options, separated by commas.
Note that $\begin{pmatrix} 1 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ are two points on this line, so a possible direction vector is
\[\begin{pmatrix} 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}.\]Then any nonzero scalar multiple of $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ can also be a direction vector.

The form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v} + t \mathbf{d}\]parameterizes a line if and only if $\mathbf{v}$ lies on the line, and $\mathbf{d}$ is a possible direction vector for the line.  Checking, we find that the possible parameterizations are $\boxed{\text{A,C}}.$
A,C
Level 3
Precalculus
Problem: The line $y = \frac{5}{3} x - \frac{17}{3}$ is to be parameterized using vectors.  Which of the following options are valid parameterizations?

(A) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} -3 \\ -5 \end{pmatrix}$

(B) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 17 \\ 5 \end{pmatrix} + t \begin{pmatrix} 6 \\ 10 \end{pmatrix}$

(C) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ -7/3 \end{pmatrix} + t \begin{pmatrix} 3/5 \\ 1 \end{pmatrix}$

(D) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 14/5 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 3/5 \end{pmatrix}$

(E) $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ -17/3 \end{pmatrix} + t \begin{pmatrix} 15 \\ -25 \end{pmatrix}$

Enter the letters of the correct options, separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine valid parameterizations of the line given by the equation y = (5/3)x - (17/3), we need to identify points on the line and direction vectors that are consistent with the slope of the line.

The slope of the line is 5/3, indicating that for every 3 units moved in the x-direction, the line moves 5 units in the y-direction. A direction vector that reflects this slope can be expressed as (3, 5) or any scalar multiple thereof, such as (6, 10) or (-3, -5).

For each option, we will check if the point lies on the line and if the direction vector is a valid representation of the slope.

Option A: The point (4, 1) is checked by substituting x = 4 into the line equation. y = (5/3)(4) - (17/3) = (20/3) - (17/3) = 1. The direction vector (-3, -5) has a slope of -5/-3 = 5/3, which is valid. Thus, A is valid.

Option B: The point (17, 5) is checked. y = (5/3)(17) - (17/3) = (85/3) - (17/3) = (68/3), which does not equal 5. The direction vector (6, 10) has a slope of 10/6 = 5/3, but since the point is not on the line, B is invalid.

Option C: The point (2, -7/3) is checked. y = (5/3)(2) - (17/3) = (10/3) - (17/3) = -7/3, confirming it lies on the line. The direction vector (3/5, 1) has a slope of 1/(3/5) = 5/3, making C valid.

Option D: The point (14/5, -1) is checked. y = (5/3)(14/5) - (17/3) = (14/3) - (17/3) = -1, confirming it lies on the line. The direction vector (1, 3/5) has a slope of (3/5)/1 = 3/5, which does not match the line's slope of 5/3. Thus, D is invalid.

Option E: The point (0, -17/3) is checked. y = (5/3)(0) - (17/3) = -17/3, confirming it lies on the line. The direction vector (15, -25) has a slope of -25/15 = -5/3, which is not consistent with the line's slope. Therefore, E is invalid.

The valid parameterizations are A and C.<|END_OF_LATENT|> The final answer is $A,C$.