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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.

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  • $\begingroup$ The centroid of a triangle is never equidistant from the vertices, unless the triangle is equilateral. The point which is equidistant from the vertices is the circumcenter. So most likely, you've misstated the problem. $\endgroup$ – quasi Oct 30 '17 at 5:35
  • $\begingroup$ @quasi That was my bad. After I came back to edit the above, I realized that I had mistakenly written that. Thanks! $\endgroup$ – Mathmath Oct 30 '17 at 5:38
  • $\begingroup$ Is there still a question? If so, please edit your posted question to correct any misstatement. $\endgroup$ – quasi Oct 30 '17 at 5:40
  • $\begingroup$ @quasi I updated the question. I took off the equidistant part and added in a midpoint. $\endgroup$ – Mathmath Oct 30 '17 at 5:46
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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*}

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  • $\begingroup$ Thank you for this. I plugged in my respective points (and I am guessing you just simply solve for x and then y or vice versa for the coordinates? Ex. m=(3,0) and p=(4,2)), so then we get $C_{x}$=3-2(4-3) which gives us a X coordinate of 5? Doing the same for y, we get y=4. Am I going about this wrong? $\endgroup$ – Mathmath Oct 30 '17 at 6:05
  • $\begingroup$ Nevermind, I see what I did wrong. Thank you for this. $\endgroup$ – Mathmath Oct 30 '17 at 6:08

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