If the middle points of the sides of a triangle be (- 2, 3), (4, - 3) and (4, 5), then the centroid of the triangle is:

Options

A. (5/3, 2)
B. (5/6, 1)
C. (2, 5/3)
D. (1, 5/6)

Answer

To find the centroid of a triangle 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:
- A = (-2, 3)
- B = (4, -3)
- C = (4, 5)

### Step 2: Use the midpoint formula
The coordinates of the vertices of the triangle can be derived from the midpoints. The midpoint formula states that if (x1, y1) and (x2, y2) are the endpoints of a segment, then the midpoint M is given by:
\[ M = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right) \]

Using this formula, we can set up equations for the vertices of the triangle.

### Step 3: Set up equations for the vertices
Let the vertices of the triangle be \( P(x_1, y_1) \), \( Q(x_2, y_2) \), and \( R(x_3, y_3) \).

1. For midpoint A (-2, 3):
\[ \frac{x_1 + x_2}{2} = -2 \implies x_1 + x_2 = -4 \quad (1) \]
\[ \frac{y_1 + y_2}{2} = 3 \implies y_1 + y_2 = 6 \quad (2) \]

2. For midpoint B (4, -3):
\[ \frac{x_2 + x_3}{2} = 4 \implies x_2 + x_3 = 8 \quad (3) \]
\[ \frac{y_2 + y_3}{2} = -3 \implies y_2 + y_3 = -6 \quad (4) \]

3. For midpoint C (4, 5):
\[ \frac{x_1 + x_3}{2} = 4 \implies x_1 + x_3 = 8 \quad (5) \]
\[ \frac{y_1 + y_3}{2} = 5 \implies y_1 + y_3 = 10 \quad (6) \]

### Step 4: Solve the equations
Now we have a system of equations to solve for \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \).

From equations (1), (3), and (5):
1. From (1): \( x_1 + x_2 = -4 \)
2. From (3): \( x_2 + x_3 = 8 \)
3. From (5): \( x_1 + x_3 = 8 \)

Adding equations (1) and (3):
\[ x_1 + x_2 + x_2 + x_3 = -4 + 8 \implies x_1 + 2x_2 + x_3 = 4 \quad (7) \]

Now adding equations (1) and (5):
\[ x_1 + x_2 + x_1 + x_3 = -4 + 8 \implies 2x_1 + x_2 + x_3 = 4 \quad (8) \]

Now we can solve for \( x_1, x_2, x_3 \) by substituting \( x_2 \) from (1) into (3) and (5).

From (1):
\[ x_2 = -4 - x_1 \]

Substituting into (3):
\[ (-4 - x_1) + x_3 = 8 \implies x_3 = 12 + x_1 \quad (9) \]

Substituting into (5):
\[ x_1 + (12 + x_1) = 8 \implies 2x_1 + 12 = 8 \implies 2x_1 = -4 \implies x_1 = -2 \]

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

Substituting \( x_1 = -2 \) into (9):
\[ x_3 = 12 + (-2) = 10 \]

Thus, we have:
- \( x_1 = -2 \)
- \( x_2 = -2 \)
- \( x_3 = 10 \)

Now, we can solve for \( y_1, y_2, y_3 \) using equations (2), (4), and (6):
1. From (2): \( y_1 + y_2 = 6 \)
2. From (4): \( y_2 + y_3 = -6 \)
3. From (6): \( y_1 + y_3 = 10 \)

Following similar steps as above, we will find:
- \( y_1 = 5 \)
- \( y_2 = 1 \)
- \( y_3 = -7 \)

### Step 5: Calculate the centroid
The coordinates of 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 values:
\[ G = \left( \frac{-2 + -2 + 10}{3}, \frac{5 + 1 + -7}{3} \right) = \left( \frac{6}{3}, \frac{-1}{3} \right) = (2, -\frac{1}{3}) \]

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