The set of vectors $\mathbf{v}$ such that
\[\operatorname{proj}_{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} \mathbf{v} = \begin{pmatrix} -\frac{5}{2} \\ -1 \end{pmatrix}\]lie on a line.  Enter the equation of this line in the form "$y = mx + b$".
So we are looking for $\begin{pmatrix} x \\ y \end{pmatrix}$ such that the projection of $\begin{pmatrix} x \\ y \end{pmatrix}$ onto $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$ is $\begin{pmatrix} -\frac{5}{2} \\ -1 \end{pmatrix}$. So the first step is to actually find the projection, which is $\frac{\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \cdot \begin{pmatrix} 5 \\ 2 \end{pmatrix}$. Which is $\frac{5x + 2y}{29} \cdot \begin{pmatrix} 5 \\ 2 \end{pmatrix}$. And we know this is equal to $\begin{pmatrix} -\frac{5}{2} \\ -1 \end{pmatrix}$, so we have the system of equations:
\begin{align*}
\frac{5x + 2y}{29} \cdot 5 &= -\frac{5}{2}, \\
\frac{5x + 2y}{29} \cdot 2 &= -1.
\end{align*} Which simplifies to
\begin{align*}
\frac{25x + 10y}{29} &= -\frac{5}{2}, \\
\frac{10x + 4y}{29} &= -1.
\end{align*} And if we multiply the second equation by 2.5, we get
\[\frac{25x + 10y}{29} = -\frac{5}{2}.\] So the two equations are the same.  This means the line is the solution to the equation
\[\frac{25x + 10y}{29} = -\frac{5}{2}.\] Which we can simplify to
\[25x + 10y = -\frac{145}{2}.\] So
\[10y = -\frac{145}{2} - 25x,\]or
\[y = -\frac{145}{20} - \frac{5}{2} x.\] This is
\[y = -\frac{29}{4} - \frac{5}{2} x.\] So the answer is $y = -\frac{5}{2} x - \frac{29}{4}$.

