If `((3)/(2),0), ((3)/(2), 6)` and `(-1, 6)` are mid-points of the sides of a triangle, then find
Centroid of the triangle

Answer

To find the centroid of the triangle given the midpoints of its sides, we can follow these steps:

### Step 1: Identify the midpoints
We are given the midpoints of the sides of the triangle:
- Midpoint D: \(\left(\frac{3}{2}, 0\right)\)
- Midpoint E: \(\left(\frac{3}{2}, 6\right)\)
- Midpoint F: \((-1, 6)\)

### Step 2: Set up equations for the vertices
Let the vertices of the triangle be \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\). The midpoints can be expressed in terms of the vertices:
1. For midpoint D:
\[ D = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) = \left(\frac{3}{2}, 0\right) \]
This gives us the equations:
\[ x_2 + x_3 = 3 \quad (1) \]
\[ y_2 + y_3 = 0 \quad (2) \]

2. For midpoint E:
\[ E = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{3}{2}, 6\right) \]
This gives us the equations:
\[ x_1 + x_3 = 3 \quad (3) \]
\[ y_1 + y_3 = 12 \quad (4) \]

3. For midpoint F:
\[ F = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (-1, 6) \]
This gives us the equations:
\[ x_1 + x_2 = -2 \quad (5) \]
\[ y_1 + y_2 = 12 \quad (6) \]

### Step 3: Solve the system of equations
Now we have a system of equations to solve for \(x_1, x_2, x_3, y_1, y_2, y_3\).

From equations (1) and (3):
- From (1): \(x_3 = 3 - x_2\)
- Substitute \(x_3\) in (3):
\[ x_1 + (3 - x_2) = 3 \implies x_1 - x_2 = 0 \implies x_1 = x_2 \]

Now substituting \(x_1\) in (5):
\[ x_1 + x_2 = -2 \implies 2x_1 = -2 \implies x_1 = -1 \implies x_2 = -1 \]
Now substituting \(x_2\) back into (1):
\[ -1 + x_3 = 3 \implies x_3 = 4 \]

Now we have \(x_1 = -1\), \(x_2 = -1\), and \(x_3 = 4\).

Next, we solve for \(y_1, y_2, y_3\) using equations (2), (4), and (6):
From (2) and (4):
- From (2): \(y_3 = -y_2\)
- Substitute \(y_3\) in (4):
\[ y_1 - y_2 = 12 \implies y_1 = y_2 + 12 \]

Now substituting \(y_1\) in (6):
\[ (y_2 + 12) + y_2 = 12 \implies 2y_2 + 12 = 12 \implies 2y_2 = 0 \implies y_2 = 0 \implies y_3 = 0 \]
Now substituting \(y_2\) back into (4):
\[ y_1 + 0 = 12 \implies y_1 = 12 \]

### Step 4: Find the centroid
Now we have the coordinates of the vertices:
- \(A(-1, 12)\)
- \(B(-1, 0)\)
- \(C(4, 0)\)

The centroid \(G\) of the triangle is given by:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the values:
\[ G = \left(\frac{-1 - 1 + 4}{3}, \frac{12 + 0 + 0}{3}\right) = \left(\frac{2}{3}, 4\right) \]

### Final Answer
The centroid of the triangle is:
\[ \boxed{\left(\frac{2}{3}, 4\right)} \]