If the coordinates of the mid points of the sides of a triangle are (1,1) ,(2,-3) and (3,4) . Find its centroid :

Options

A. `(3 , 2/3)`
B. `(2, 3/4)`
C. `(2, 2/3)`
D. none of these

Answer

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

### Step 1: Identify the midpoints
The midpoints of the sides of the triangle are given as:
- M1 = (1, 1)
- M2 = (2, -3)
- M3 = (3, 4)

### Step 2: Use the midpoint formula to find the vertices
Let the vertices of the triangle be A(x1, y1), B(x2, y2), and C(x3, y3). The midpoints can be expressed in terms of the vertices as follows:

1. Midpoint M1 between A and B:
\[ M1 = \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) = (1, 1) \]

2. Midpoint M2 between B and C:
\[ M2 = \left(\frac{x2 + x3}{2}, \frac{y2 + y3}{2}\right) = (2, -3) \]

3. Midpoint M3 between C and A:
\[ M3 = \left(\frac{x3 + x1}{2}, \frac{y3 + y1}{2}\right) = (3, 4) \]

### Step 3: Set up equations from the midpoints
From the midpoint equations, we can set up the following system of equations:

1. From M1:
\[ \frac{x1 + x2}{2} = 1 \quad \Rightarrow \quad x1 + x2 = 2 \quad (1) \]
\[ \frac{y1 + y2}{2} = 1 \quad \Rightarrow \quad y1 + y2 = 2 \quad (2) \]

2. From M2:
\[ \frac{x2 + x3}{2} = 2 \quad \Rightarrow \quad x2 + x3 = 4 \quad (3) \]
\[ \frac{y2 + y3}{2} = -3 \quad \Rightarrow \quad y2 + y3 = -6 \quad (4) \]

3. From M3:
\[ \frac{x3 + x1}{2} = 3 \quad \Rightarrow \quad x3 + x1 = 6 \quad (5) \]
\[ \frac{y3 + y1}{2} = 4 \quad \Rightarrow \quad y3 + y1 = 8 \quad (6) \]

### Step 4: Solve the system of equations
Now we have a system of 6 equations. We can solve for x1, x2, x3, y1, y2, and y3.

From equations (1), (3), and (5):
1. From (1): \( x2 = 2 - x1 \)
2. Substitute \( x2 \) into (3):
\[ (2 - x1) + x3 = 4 \quad \Rightarrow \quad x3 = 2 + x1 \quad (7) \]
3. Substitute \( x3 \) into (5):
\[ (2 + x1) + x1 = 6 \quad \Rightarrow \quad 2x1 + 2 = 6 \quad \Rightarrow \quad 2x1 = 4 \quad \Rightarrow \quad x1 = 2 \]
4. Substitute \( x1 \) back to find \( x2 \) and \( x3 \):
\[ x2 = 2 - 2 = 0 \]
\[ x3 = 2 + 2 = 4 \]

Now we have \( x1 = 2, x2 = 0, x3 = 4 \).

Next, solve for y-coordinates using equations (2), (4), and (6):
1. From (2): \( y2 = 2 - y1 \)
2. Substitute \( y2 \) into (4):
\[ (2 - y1) + y3 = -6 \quad \Rightarrow \quad y3 = -8 + y1 \quad (8) \]
3. Substitute \( y3 \) into (6):
\[ (-8 + y1) + y1 = 8 \quad \Rightarrow \quad 2y1 - 8 = 8 \quad \Rightarrow \quad 2y1 = 16 \quad \Rightarrow \quad y1 = 8 \]
4. Substitute \( y1 \) back to find \( y2 \) and \( y3 \):
\[ y2 = 2 - 8 = -6 \]
\[ y3 = -8 + 8 = 0 \]

Now we have the vertices:
- A(2, 8)
- B(0, -6)
- C(4, 0)

### Step 5: Calculate the centroid
The formula for the centroid (G) of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) is:
\[ G = \left(\frac{x1 + x2 + x3}{3}, \frac{y1 + y2 + y3}{3}\right) \]
Substituting the values:
\[ G = \left(\frac{2 + 0 + 4}{3}, \frac{8 + (-6) + 0}{3}\right) = \left(\frac{6}{3}, \frac{2}{3}\right) = (2, \frac{2}{3}) \]

### Final Answer
The centroid of the triangle is \( G(2, \frac{2}{3}) \).

---