Finding third vertex given 2 points and centroid I am given the following two vertices: $A=(0,0)$, $B=(6,0)$ and centroid at $P=(4,2)$. Instead of finding midpoints such as $\overline{AC}$ and $\overline{BC}$ (because I end up getting really confused), I used the formula for the centroid, which is:
$$\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}$$
Note, I am not given any information regarding if the triangle is a right triangle, isosceles triangle etc.
Plugging in our given values into our centroid equation, we get:
$$\frac{0+6+x}{3},\frac{0+0+y}{3}$$ or
$$\frac{6+x}{3},\frac{y}{3}$$
If we set each respective x,y value equal to our centroid points, we get:
$$\frac{6+x}{3}=4,\frac{y}{3}=2$$
and we can clearly see that our vertex $C=(6,6)$. I have triple checked and got the same point over and over again and I am pretty sure this is correct.  So,
A) Is my vertex C correct? (Note I am just making sure I am doing this properly)
B) How does one go about properly solving this by using the method of first finding the midpoint of $\overline{BC}$?
Note, the midpoint I get for $\overline{BC}$ is $$\frac{6+x}{2},\frac{y}{2}$$
I do not know what other bisector I need. 
EDIT: 
I figured I would need the midpoint for $\overline{AC}$ as well which is $$\frac{0+x}{2},\frac{0+y}{2}$$ or $$\frac{x}{2},\frac{y}{2}$$. I do not know where to go from here. I will attempt to find slope of each and find the equation for each given point.

Your method is fine$\,-\,$there no need to do it another way.

But to satisfy your curiosity, here is one such other way . . .

Find the midpoint, $M$ say, of $AB$. 

Since the centroid $P$ divides the median $CM$ into two parts with length ratio $CP:\!PM=2\!:\!1$, it follows that $C = P+2(P-M)$.

Then, since $M={\large{\frac{A+B}{2}}}=(3,0),\;$we get
\begin{align*}
C &= P + 2(P-M)\\[4pt]
&=(4,2)+2\bigl((4,2)-(3,0)\bigr)\\[4pt]
&=(4,2)+2(1,2)\\[4pt]
&=(4,2)+(2,4)\\[4pt]
&=(6,6)\\[4pt]
\end{align*} 
