The line $y = \frac{1}{2} x + 4$ is paramterized by the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -7 \\ s \end{pmatrix} + t \begin{pmatrix} l \\ -5 \end{pmatrix}.\]Enter the ordered pair $(s,l).$
Taking $t = 0,$ we find $\begin{pmatrix} -7 \\ s \end{pmatrix}$ lies on the line.  Then
\[s = \frac{1}{2} (-7) + 4 = \frac{1}{2}.\]Taking $t = 1,$ we get
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -7 \\ 1/2 \end{pmatrix} + \begin{pmatrix} l \\ -5 \end{pmatrix} = \begin{pmatrix} -7 + l \\ -9/2 \end{pmatrix}.\]Then
\[-\frac{9}{2} = \frac{1}{2} (-7 + l) + 4.\]Solving for $l,$ we find $l = -10.$

Hence, $(r,k) = \boxed{\left( \frac{1}{2}, -10 \right)}.$
\le(\frac{1}{2},-10\right)
Level 4
Precalculus
Problem: The line $y = \frac{1}{2} x + 4$ is paramterized by the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -7 \\ s \end{pmatrix} + t \begin{pmatrix} l \\ -5 \end{pmatrix}.\]Enter the ordered pair $(s,l).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ordered pair $(s, l)$, we need to ensure that the parameterization of the line matches the given line equation. The line equation is in slope-intercept form, where the slope is $\frac{1}{2}$ and the y-intercept is $4$. 

First, we identify the direction vector of the line. The slope $\frac{1}{2}$ indicates that for every increase of $1$ in $x$, $y$ increases by $\frac{1}{2}$. Therefore, a suitable direction vector can be expressed as $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, which maintains the ratio of the rise over run.

Next, we need to express this direction vector in a form that matches the parameterization given in the problem. The parameterization uses the vector $\begin{pmatrix} l \\ -5 \end{pmatrix}$. To maintain the same direction, we can set up a proportionality condition. The direction vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ must be a scalar multiple of $\begin{pmatrix} l \\ -5 \end{pmatrix}$.

This gives us the equations:
1. $\frac{2}{l} = \frac{1}{-5}$.

Cross-multiplying yields:
2. $2 \cdot (-5) = 1 \cdot l$, which simplifies to $l = -10$.

Now, we need to find the value of $s$. The point $\begin{pmatrix} -7 \\ s \end{pmatrix}$ must lie on the line defined by the equation $y = \frac{1}{2} x + 4$. We substitute $x = -7$ into the line equation to find the corresponding $y$ value:
3. $y = \frac{1}{2}(-7) + 4 = -\frac{7}{2} + 4 = -\frac{7}{2} + \frac{8}{2} = \frac{1}{2}$.

Thus, we find that $s = \frac{1}{2}$. 

Combining these results, we conclude that the ordered pair is $(s, l) = \left(\frac{1}{2}, -10\right)$.<|END_OF_LATENT|> The final answer is $\le(\frac{1}{2},-10\right)$.