The line $y = \frac{-12x + 74}{5}$ is parameterized in the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{v} + t \mathbf{d},\]so that for $x \le 7,$ the distance between $\begin{pmatrix} x \\ y \end{pmatrix}$ and $\begin{pmatrix} 7 \\ -2 \end{pmatrix}$ 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 $\begin{pmatrix} 7 \\ -2 \end{pmatrix}$ is $t = 0,$ so $\mathbf{v} = \begin{pmatrix} 7 \\ -2 \end{pmatrix}.$  Thus,
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \end{pmatrix} + t \mathbf{d}.\]Then for $x \le 7,$
\[\left\| \begin{pmatrix} x - 7 \\ y + 2 \end{pmatrix} \right\| = \left\| \begin{pmatrix} x - 7 \\ \frac{-12x + 84}{5} \end{pmatrix} \right\| = \left\| \begin{pmatrix} 1 \\ -\frac{12}{5} \end{pmatrix} \right\| (7 - x) = \frac{13}{5} (7 - x).\]We want this to be $t,$ so $t = \frac{13}{5} (7 - x).$  Then $x = 7 - \frac{5}{13} t,$ and $y = \frac{-12x + 74}{5} = \frac{12}{13} t - 2,$ so
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 - \frac{5}{13} t \\ \frac{12}{13} t - 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \end{pmatrix} + t \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}.\]Thus, $\mathbf{d} = \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}.$