The co- ordinates of vertex A of `DeltaABC` are `(-4, 2)` and a point D which is mid point of BC are (2, 5) . The coordinates of centroid of `Delta ABC` are

Options

A. (0, 4)
B. `(-1, (7)/(2))`
C. `(-2, (7)/(3))`
D. `(0, 2)`

Answer

To find the coordinates of the centroid of triangle ABC, we can follow these steps:

### Step 1: Identify the given coordinates
The coordinates of vertex A are given as:
- \( A(-4, 2) \)

The coordinates of point D, which is the midpoint of line segment BC, are given as:
- \( D(2, 5) \)

### Step 2: Set up the equations for the midpoint D
Since D is the midpoint of segment BC, we can express the coordinates of points B and C as:
- \( B(x_1, y_1) \)
- \( C(x_2, y_2) \)

The formula for the midpoint D is:
\[ D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Given that \( D(2, 5) \), we can set up the following equations:
1. \( \frac{x_1 + x_2}{2} = 2 \)
2. \( \frac{y_1 + y_2}{2} = 5 \)

### Step 3: Solve for \( x_1 + x_2 \) and \( y_1 + y_2 \)
From the first equation:
\[ x_1 + x_2 = 2 \times 2 = 4 \]

From the second equation:
\[ y_1 + y_2 = 2 \times 5 = 10 \]

### Step 4: Use the coordinates to find the centroid
The formula for the centroid \( G \) of triangle ABC is given by:
\[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) \]
Substituting the known values:
- \( x_A = -4 \)
- \( y_A = 2 \)
- \( x_B + x_C = x_1 + x_2 = 4 \)
- \( y_B + y_C = y_1 + y_2 = 10 \)

Thus, we can write:
\[ G_x = \frac{-4 + 4}{3} = \frac{0}{3} = 0 \]
\[ G_y = \frac{2 + 10}{3} = \frac{12}{3} = 4 \]

### Step 5: Final coordinates of the centroid
The coordinates of the centroid \( G \) are:
\[ G(0, 4) \]