Answer
Two possible third vertices: $(\pm2\sqrt 3, 2)$
Work Step by Step
Step 1. Let $A(0,4), B(0,0)$ be the two given vertices and $C(x,y)$ be the third vertex, we have $s=AB=AC=BC$ or $\sqrt {x^2+(y-4)^2}=\sqrt {x^2+y^2}=4$
Step 2. From the first equal sign, we have $y-4=\pm y$ which leads to $y=2$
Step 3. Thus $x^2+(2)^2=4^2$ or $x=\pm2\sqrt 3$ Step 4. There are two possible third vertices: $(\pm2\sqrt 3, 2)$
