Okay, so I need to find the equation of the line on which all vectors v lie, given that their projection onto the vector (2, 1) is the zero vector. Hmm, let's start by recalling what a projection of a vector onto another vector means.

The projection of vector v onto another vector u is given by the formula:

proj_u(v) = ( (v · u) / ||u||² ) * u

So, if the projection is the zero vector, that means (v · u) / ||u||² * u = 0. Since u is (2, 1), which is not the zero vector, the only way this can happen is if the dot product v · u is zero. Wait, is that right? Because if the dot product is zero, then the projection would be zero vector. So, yeah, that makes sense.

Therefore, the condition here is that the dot product of v with u must be zero. Let me write that down. Let’s denote v as (x, y). Then, the dot product v · u is 2x + y. So, 2x + y = 0. That's the equation of a line. So, the line is 2x + y = 0. But the problem asks for it in the form y = mx + b. So, solving for y, we get y = -2x + 0. Therefore, the equation is y = -2x. 

Wait, that seems straightforward. Let me verify. If the projection of v onto u is zero, then v is orthogonal to u. Because projection is zero means that the component of v in the direction of u is zero, which implies that v is perpendicular to u. So, the set of all such vectors v lie on the line that is perpendicular to u.

Since u is (2, 1), the direction vector of the line we are looking for is perpendicular to (2, 1). The direction vector perpendicular to (2, 1) is (-1, 2) or (1, -2). But to find the equation of the line, we need a point and the slope. Wait, but since the line passes through the origin? Wait, no. Wait, the projection is zero only when v is orthogonal to u, so the line is the set of all vectors orthogonal to u. So, the line is the set of vectors v such that v is perpendicular to u. So, the line passes through the origin because if v is the zero vector, it's trivially orthogonal to u. So, the line passes through (0, 0) and has a slope that is the negative reciprocal of the slope of u. 

Wait, the slope of u is 1/2 (since u is (2,1), so rise over run is 1/2). Therefore, the slope of the perpendicular line is -2. Hence, the equation is y = -2x. Which matches what I found earlier. So, that seems correct. Let me check with a point. Let’s take a point on the line y = -2x. For example, (1, -2). The projection of (1, -2) onto (2,1) should be zero. Let's compute it. The dot product is 2*1 + 1*(-2) = 2 -2 = 0. So, the projection is (0,0). Perfect. Another point: (0,0). Projection is zero. Another point: (2, -4). Dot product is 2*2 + 1*(-4) = 4 -4 = 0. Also works. So, yes, the line y = -2x is correct.

Therefore, the answer should be y = -2x. So, the equation is y = -2x. I think that's it. Let me just recap. The projection being zero implies orthogonality, which gives the equation of the line as y = -2x. Yep, that seems right.
Therefore, the equation of the line on which all such vectors $\mathbf{v}$ lie is $\boxed{y = -2x}$.
</think>