The coordinates of the middle points of the sides of a triangle are (4, 2), (3, 3) and (2, 2), then coordinates of centroid are

Options

A. (3, 7/3)
B. (3, 3)
C. (4, 3)
D. (3, 4)

Answer

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

### Step-by-Step Solution:

1. **Identify the Midpoints**:
The midpoints of the sides of the triangle are given as:
- \( D(4, 2) \)
- \( E(3, 3) \)
- \( F(2, 2) \)

2. **Set Up the Equations**:
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:
- For midpoint \( D \):
\[ D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (4, 2) \]
This gives us two equations:
\[ \frac{x_1 + x_2}{2} = 4 \quad \Rightarrow \quad x_1 + x_2 = 8 \quad \text{(1)} \]
\[ \frac{y_1 + y_2}{2} = 2 \quad \Rightarrow \quad y_1 + y_2 = 4 \quad \text{(2)} \]

- For midpoint \( E \):
\[ E = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (3, 3) \]
This gives us:
\[ \frac{x_2 + x_3}{2} = 3 \quad \Rightarrow \quad x_2 + x_3 = 6 \quad \text{(3)} \]
\[ \frac{y_2 + y_3}{2} = 3 \quad \Rightarrow \quad y_2 + y_3 = 6 \quad \text{(4)} \]

- For midpoint \( F \):
\[ F = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (2, 2) \]
This gives us:
\[ \frac{x_1 + x_3}{2} = 2 \quad \Rightarrow \quad x_1 + x_3 = 4 \quad \text{(5)} \]
\[ \frac{y_1 + y_3}{2} = 2 \quad \Rightarrow \quad y_1 + y_3 = 4 \quad \text{(6)} \]

3. **Solve the System of Equations**:
Now we have a system of equations for \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \):
- From equations (1), (3), and (5):
\[ x_1 + x_2 = 8 \quad (1) \]
\[ x_2 + x_3 = 6 \quad (3) \]
\[ x_1 + x_3 = 4 \quad (5) \]

Adding equations (1) and (3):
\[ (x_1 + x_2) + (x_2 + x_3) = 8 + 6 \quad \Rightarrow \quad x_1 + 2x_2 + x_3 = 14 \quad \text{(7)} \]

Now subtract equation (5) from (7):
\[ (x_1 + 2x_2 + x_3) - (x_1 + x_3) = 14 - 4 \quad \Rightarrow \quad 2x_2 = 10 \quad \Rightarrow \quad x_2 = 5 \]

Substitute \( x_2 = 5 \) back into equations (1) and (3):
- From (1):
\[ x_1 + 5 = 8 \quad \Rightarrow \quad x_1 = 3 \]
- From (3):
\[ 5 + x_3 = 6 \quad \Rightarrow \quad x_3 = 1 \]

Thus, we have:
\[ x_1 = 3, \quad x_2 = 5, \quad x_3 = 1 \]

Now, we can solve for \( y_1, y_2, y_3 \) using equations (2), (4), and (6):
- From (2):
\[ y_1 + y_2 = 4 \quad (2) \]
- From (4):
\[ y_2 + y_3 = 6 \quad (4) \]
- From (6):
\[ y_1 + y_3 = 4 \quad (6) \]

Adding equations (2) and (4):
\[ (y_1 + y_2) + (y_2 + y_3) = 4 + 6 \quad \Rightarrow \quad y_1 + 2y_2 + y_3 = 10 \quad \text{(8)} \]

Now subtract equation (6) from (8):
\[ (y_1 + 2y_2 + y_3) - (y_1 + y_3) = 10 - 4 \quad \Rightarrow \quad 2y_2 = 6 \quad \Rightarrow \quad y_2 = 3 \]

Substitute \( y_2 = 3 \) back into equations (2) and (4):
- From (2):
\[ y_1 + 3 = 4 \quad \Rightarrow \quad y_1 = 1 \]
- From (4):
\[ 3 + y_3 = 6 \quad \Rightarrow \quad y_3 = 3 \]

Thus, we have:
\[ y_1 = 1, \quad y_2 = 3, \quad y_3 = 3 \]

4. **Calculate the Centroid**:
The coordinates of the centroid \( G \) of the triangle are given by:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the values:
\[ G\left( \frac{3 + 5 + 1}{3}, \frac{1 + 3 + 3}{3} \right) = G\left( \frac{9}{3}, \frac{7}{3} \right) = G(3, \frac{7}{3}) \]

### Final Answer:
The coordinates of the centroid are \( (3, \frac{7}{3}) \).