The line passes through $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3 \end{pmatrix},$ so its direction vector is proportional to
\[\begin{pmatrix} -1 \\ 3 \end{pmatrix} - \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}.\]To get an $x$-coordinate of $-7,$ we can multiply this vector by the scalar $-\frac{7}{4}.$  This gives us
\[-\frac{7}{4} \begin{pmatrix} 4 \\ -1 \end{pmatrix} = \begin{pmatrix} -7 \\ 7/4 \end{pmatrix}.\]Therefore, $b = \boxed{\frac{7}{4}}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
The following line is parameterized, so that its direction vector is of the form $\begin{pmatrix} -7 \\ b \end{pmatrix}.$  Find $b.$

[asy]
unitsize(0.4 cm);

pair A, B, L, R;
int i, n;

for (i = -8; i <= 8; ++i) {
  draw((i,-8)--(i,8),gray(0.7));
  draw((-8,i)--(8,i),gray(0.7));
}

draw((-8,0)--(8,0),Arrows(6));
draw((0,-8)--(0,8),Arrows(6));

A = (-5,4);
B = (-1,3);
L = extension(A, B, (-8,0), (-8,1));
R = extension(A, B, (8,0), (8,1));

draw(L--R, red);

label("$x$", (8,0), E);
label("$y$", (0,8), N);
[/asy]
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\frac{7}{4}
The following line is parameterized, so that its direction vector is of the form $\begin{pmatrix} -7 \\ b \end{pmatrix}.$  Find $b.$

[asy]
unitsize(0.4 cm);

pair A, B, L, R;
int i, n;

for (i = -8; i <= 8; ++i) {
  draw((i,-8)--(i,8),gray(0.7));
  draw((-8,i)--(8,i),gray(0.7));
}

draw((-8,0)--(8,0),Arrows(6));
draw((0,-8)--(0,8),Arrows(6));

A = (-5,4);
B = (-1,3);
L = extension(A, B, (-8,0), (-8,1));
R = extension(A, B, (8,0), (8,1));

draw(L--R, red);

label("$x$", (8,0), E);
label("$y$", (0,8), N);
[/asy]
test