A triangle having a vertex as `(1,2)` has mid-point of sides passig from this vertex as `(-1,1)` and `(2,3)` thent he centroid of the triangle is

Options

A. `(1/3,5/3)`
B. `(1/3,2)`
C. `(1/7/3)`
D. `(1/3,1)`

Explanation

To find the centroid of the triangle with a vertex at \( (1, 2) \) and midpoints of the sides passing through this vertex at \( (-1, 1) \) and \( (2, 3) \), we will follow these steps:

### Step 1: Identify the given points
- Vertex \( A = (1, 2) \)
- Midpoint \( M_1 = (-1, 1) \) for side \( AB \)
- Midpoint \( M_2 = (2, 3) \) for side \( AC \)

### Step 2: Use the midpoint formula to find coordinates of points \( B \) and \( C \)
The midpoint formula states that the midpoint \( M \) of a line segment joining points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

#### Finding point \( B \):
For midpoint \( M_1 = (-1, 1) \):
\[ \left( \frac{1 + x_2}{2}, \frac{2 + y_2}{2} \right) = (-1, 1) \]
This gives us two equations:
1. \( \frac{1 + x_2}{2} = -1 \)
2. \( \frac{2 + y_2}{2} = 1 \)

From the first equation:
\[ 1 + x_2 = -2 \implies x_2 = -3 \]

From the second equation:
\[ 2 + y_2 = 2 \implies y_2 = 0 \]

Thus, point \( B = (-3, 0) \).

#### Finding point \( C \):
For midpoint \( M_2 = (2, 3) \):
\[ \left( \frac{1 + x_3}{2}, \frac{2 + y_3}{2} \right) = (2, 3) \]
This gives us two equations:
1. \( \frac{1 + x_3}{2} = 2 \)
2. \( \frac{2 + y_3}{2} = 3 \)

From the first equation:
\[ 1 + x_3 = 4 \implies x_3 = 3 \]

From the second equation:
\[ 2 + y_3 = 6 \implies y_3 = 4 \]

Thus, point \( C = (3, 4) \).

### Step 3: Calculate the centroid of triangle \( ABC \)
The centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Substituting the coordinates:
- \( A = (1, 2) \)
- \( B = (-3, 0) \)
- \( C = (3, 4) \)

Calculating the centroid:
\[ G_x = \frac{1 + (-3) + 3}{3} = \frac{1}{3} \]
\[ G_y = \frac{2 + 0 + 4}{3} = \frac{6}{3} = 2 \]

Thus, the coordinates of the centroid \( G \) are:
\[ G = \left( \frac{1}{3}, 2 \right) \]

### Final Answer:
The centroid of the triangle is \( \left( \frac{1}{3}, 2 \right) \).