To find the coordinates of vertex [tex]\( C \)[/tex] given that [tex]\( A (2, -2) \)[/tex] and [tex]\( B (-2, 1) \)[/tex] are vertices and the centroid of triangle [tex]\( ABC \)[/tex] is [tex]\(\left(\frac{5}{3}, \frac{1}{3}\right)\)[/tex], we proceed as follows:
### Step 1: Finding the coordinates of [tex]\( C \)[/tex]
The coordinates of the centroid [tex]\( G \)[/tex] of a triangle with vertices [tex]\( A (x_1, y_1) \)[/tex], [tex]\( B (x_2, y_2) \)[/tex], and [tex]\( C (x_3, y_3) \)[/tex] are given by:
[tex]\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\][/tex]
Given:
[tex]\[
G = \left( \frac{5}{3}, \frac{1}{3} \right)
\][/tex]
[tex]\[
A = (2, -2), \quad B = (-2, 1)
\][/tex]
Let's denote [tex]\( C \)[/tex] as [tex]\( (x_3, y_3) \)[/tex]. Plugging in the known values into the centroid formula:
[tex]\[
\frac{x_1 + x_2 + x_3}{3} = \frac{5}{3}
\][/tex]
[tex]\[
\frac{y_1 + y_2 + y_3}{3} = \frac{1}{3}
\][/tex]
Solving for [tex]\( x_3 \)[/tex]:
[tex]\[
\frac{2 + (-2) + x_3}{3} = \frac{5}{3}
\][/tex]
[tex]\[
\frac{x_3}{3} = \frac{5}{3}
\][/tex]
[tex]\[
x_3 = 5
\][/tex]
Solving for [tex]\( y_3 \)[/tex]:
[tex]\[
\frac{-2 + 1 + y_3}{3} = \frac{1}{3}
\][/tex]
[tex]\[
\frac{-1 + y_3}{3} = \frac{1}{3}
\][/tex]
[tex]\[
-1 + y_3 = 1
\][/tex]
[tex]\[
y_3 = 2
\][/tex]
So, the coordinates of [tex]\( C \)[/tex] are [tex]\( (5, 2) \)[/tex].
### Step 2: Finding the coordinates of the circumcenter
To find the coordinates of the circumcenter, we need the perpendicular bisectors of at least two sides of the triangle.
#### Side AB
The midpoint of [tex]\( AB = \left( \frac{2 + (-2)}{2}, \frac{-2 + 1}{2} \right) = (0, -0.5) \)[/tex].
The slope of [tex]\( AB \)[/tex] is [tex]\( m_{AB} = \frac{1 - (-2)}{-2 - 2} = \frac{3}{-4} = -\frac{3}{4} \)[/tex].
The slope of the perpendicular bisector of [tex]\( AB \)[/tex] is the negative reciprocal, [tex]\( m_{\perp AB} = \frac{4}{3} \)[/tex].
#### Side BC
The midpoint of [tex]\( BC = \left( \frac{-2 + 5}{2}, \frac{1 + 2}{2} \right) = \left( \frac{3}{2}, \frac{3}{2} \right) \)[/tex].
The slope of [tex]\( BC \)[/tex] is [tex]\( m_{BC} = \frac{2 - 1}{5 - (-2)} = \frac{1}{7} \)[/tex].
The slope of the perpendicular bisector of [tex]\( BC \)[/tex] is the negative reciprocal, [tex]\( m_{\perp BC} = -7 \)[/tex].
We solve the equations of these bisectors to find the circumcenter.
The equation of the perpendicular bisector of [tex]\( AB \)[/tex] is:
[tex]\[
y + 0.5 = \frac{4}{3}(x - 0)
\][/tex]
[tex]\[
y = \frac{4}{3}x - 0.5
\][/tex]
The equation of the perpendicular bisector of [tex]\( BC \)[/tex] is:
[tex]\[
y - \frac{3}{2} = -7 \left( x - \frac{3}{2} \right)
\][/tex]
[tex]\[
y = -7x + \frac{21}{2} + \frac{3}{2}
\][/tex]
[tex]\[
y = -7x + 12
\][/tex]
Setting the equations equal to find their intersection:
[tex]\[
\frac{4}{3}x - 0.5 = -7x + 12
\][/tex]
[tex]\[
\frac{4}{3}x + 7x = 12.5
\][/tex]
[tex]\[
\frac{4x + 21x}{3} = 12.5
\][/tex]
[tex]\[
25x = 37.5
\][/tex]
[tex]\[
x = 1.5
\][/tex]
To find [tex]\( y \)[/tex]:
[tex]\[
y = \frac{4}{3}(1.5) - 0.5 = 2 - 0.5 = 1.5
\][/tex]
Thus, the coordinates of the circumcenter are [tex]\( (1.5, 1.5) \)[/tex].
In conclusion, the coordinates of vertex [tex]\( C \)[/tex] are [tex]\( (5, 2) \)[/tex] and the coordinates of the circumcenter are [tex]\( (1.5, 1.5) \)[/tex].
