Answer
There are $\textbf{two}$ possible vertices in the triangle: $(2\sqrt{3},2)$ and $(-2\sqrt{3}, 2)$
Work Step by Step
Two vertices of the triangle is on $y$ axis. Thus the $y$ coordinate on the third vertex must be in midway of $0$ and $4$ such that $y=2$.
Third vertex $(\pm x,2)$
The triangle $(0,0), (0,2),$ and $(\pm,2)$ is a $30-60-90$ triangle.
Hypotenuse $=4$
Shortest leg $=2$
Middle leg $= ?$
We can use pythagorean theorem.
Middle leg $=\sqrt{4^2 -2^2}$
$\qquad \qquad \enspace= \pm \sqrt{12}$
$\qquad \qquad \enspace= \pm 2\sqrt{3}$
Therefore, there are $\textbf{two}$ possible vertices in the triangle: $(2\sqrt{3},2)$ and $(-2\sqrt{3}, 2)$
