The line $y = \frac{3x - 5}{4}$ is parameterized in the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v} + t \mathbf{d},\]so that for $x \ge 3,$ the distance between $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ is $t.$  Find $\mathbf{d}.$
Setting $t = 0,$ we get
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v}.\]But the distance between $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ is $t = 0,$ so $\mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}.$  Thus,
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + t \mathbf{d}.\]Then for $x \ge 3,$
\[\left\| \begin{pmatrix} x - 3 \\ y - 1 \end{pmatrix} \right\| = \left\| \begin{pmatrix} x - 3 \\ \frac{3x - 9}{4} \end{pmatrix} \right\| = \left\| \begin{pmatrix} 1 \\ \frac{3}{4} \end{pmatrix} \right\| (x - 3) = \frac{5}{4} (x - 3).\]We want this to be $t,$ so $t = \frac{5}{4} (x - 3).$  Then $x = \frac{4}{5} t + 3,$ and $y = \frac{3x - 5}{4} = \frac{3}{5} t + 1,$ so
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{4}{5} t + 3 \\ \frac{3}{5} t + 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + t \begin{pmatrix} 4/5 \\ 3/5 \end{pmatrix}.\]Thus, $\mathbf{d} = \boxed{\begin{pmatrix} 4/5 \\ 3/5 \end{pmatrix}}.$
\begin{pmatrix}4/5\3/5\end{pmatrix}
Level 5
Precalculus
Problem: The line $y = \frac{3x - 5}{4}$ is parameterized in the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v} + t \mathbf{d},\]so that for $x \ge 3,$ the distance between $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ is $t.$  Find $\mathbf{d}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the direction vector \(\mathbf{d}\), we start by recognizing that the line is given in slope-intercept form. The slope of the line \(y = \frac{3x - 5}{4}\) is \(\frac{3}{4}\). This indicates that for every increase of 4 in \(x\), \(y\) increases by 3. 

Next, we need to express the direction vector \(\mathbf{d}\) in terms of its components. The direction vector can be represented as \(\begin{pmatrix} a \\ b \end{pmatrix}\), where \(a\) and \(b\) correspond to the changes in \(x\) and \(y\) respectively. Given the slope, we can derive the relationship between \(a\) and \(b\) as follows:

\[
\frac{b}{a} = \frac{3}{4} \implies b = \frac{3}{4}a.
\]

To simplify calculations, we can choose \(a = 4\) (a convenient choice that maintains the ratio). Substituting this value gives:

\[
b = \frac{3}{4} \cdot 4 = 3.
\]

Thus, one possible direction vector is \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\). However, we need to ensure that \(\mathbf{d}\) is a unit vector, as the distance \(t\) is defined in terms of a unit direction. 

To convert \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\) into a unit vector, we calculate its magnitude:

\[
\|\begin{pmatrix} 4 \\ 3 \end{pmatrix}\| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.
\]

Now, we divide each component of the vector by its magnitude to obtain the unit vector:

\[
\mathbf{d} = \begin{pmatrix} \frac{4}{5} \\ \frac{3}{5} \end{pmatrix}.
\]

This unit vector \(\mathbf{d}\) maintains the direction of the line while ensuring that the distance \(t\) is measured correctly. Thus, the final answer is \(\begin{pmatrix} \frac{4}{5} \\ \frac{3}{5} \end{pmatrix}\).<|END_OF_LATENT|> The final answer is $\begin{pmatrix}4/5\3/5\end{pmatrix}$.