The centroid of triange $ABC$ is located at $P(14,14)$ with points $A(2,12)$ and $B(22,6)$
What are the coordinates of vertex $C$?
Explain how you found the answer.
I've gotten a question of the sort one too many times without being able to find a concrete way of getting a solution. Anyone have an idea?
I supposed I would solve it like this:
Take the formula to find a centroid then plug all the values in.
Solve for the missing value.
Yet I'm unsure if that would contrain the formula enough that I would come to a solution without two variables as I need an $x$ and a $y$ of vertex $C$. Some sort of system of equation would need to be in the works right?
I can only guess what you are asking. I, therefore, explain the whole thing in 2 steps.
I. Find the co-ordinates of the centroid, P(x, y) for ⊿ABC where $A= (x_1, y_1), B= (x_2, y_2)$ and $C=(x_3, y_3)$.
1.1 Let $D(p,q)$ be the midpoint of BC. Then, by mid-pt formula, $p = \frac {x_2 + x_3}{ 2}$, (similar result for q).
1.2 From geometrical knowledge, P(x, y) must lie on the median AD and, AP : PD = 2 : 1.
1.3 Apply the section formula to get $x = \frac {x_1 + x_2 + x_3} {3}$, (similar result for y).
II. Find $x_3$ if $x_1, x_2$ and $x$ are known.
2.1 Just plug in the given data at the right places in the developed formula, $x_3$ can then be found.