Conjecture
For $x+y=2p, p \in \mathbb{P}$, there are visible points $v=p-1$.
Example
Let $p=7$, then $v=6$.
Similarly for
$p=11 \rightarrow v=10$
$p=13 \rightarrow v=12$
$p=17 \rightarrow v=16$
...
Question
How to prove the conjecture for any $p$?
I assume you are only interested in points in the positive quadrant of the plane. Recall that a point $(a, b)$ is visible from the origin if and only if $\mathrm{gcd}(a, b) = 1$.
Now, the points that lie on the line $y = -x + 2p$ in the positive quadrant are of the form $(i, 2p - i)$ for $i \in \{0, \ldots, 2p\}$.
If $i$ is even, then $\mathrm{gcd}(i, 2p - i) \geq 2$, hence these points are not visible from the origin.
If $i$ is odd, we show that the only point that is not visible from the origin is the point $(p, p)$. Assume that $\mathrm{gcd}(i, 2p - i) = d > 1$, then we have $i = da$, $2p - i = db$ for some integers $a, b$. Since $i$ is odd, $d$ must odd as well. Since $2p = d(a + b)$ you have that $d \mid 2p$, so we must have $d \mid p$. Hence $d = p$. Thus, the only point $(i, 2p - i)$ with $i$ odd that is not visible from the origin is the point $(p, 2p - i) = (p, p)$.
Thus, since there are $p - 1$ odd numbers in the interval $[0, 2p]$ that are not equal to $p$, we conclude there are $p - 1$ visible points.