Let $ \mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} $. From the formula of a projection, \begin{align*} \operatorname{proj}_{\begin{pmatrix} 2 \\ 1 \end{pmatrix}} \mathbf{v} &= \frac{\mathbf{v} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ 1 \end{pmatrix} \right\|^2} \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \frac{\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix}}{5} \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \frac{2x + y}{5} \begin{pmatrix} 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}.\end{align*}Then \[\frac{2x + y}{5} = 0,\]so $ 2x + y = 0 $. Thus, the equation of the line is $ \boxed{y = -2x} $.