The set of vectors $\mathbf{v}$ such that
\[\operatorname{proj}_{\begin{pmatrix} 2 \\ 1 \end{pmatrix}} \mathbf{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]lie on a line.  Enter the equation of this line in the form "$y = mx + b$".
I want to find the set of vectors $\mathbf{v}$ that have no component in the direction of $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. This means that $\mathbf{v}$ must be orthogonal to $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. To check orthogonality, I can use the dot product: $\mathbf{v} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix} = 0$. Writing $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$, this gives me the equation $2x + y = 0$. This is the equation of a line with slope $-2$ and $y$-intercept $0$. Therefore, the answer is

