If `A(1, 4, 2), B(-2, 3, -5)` are two vertices A and B and `G((4)/(3), 0, (-2)/(3))` is the centroid of the `triangleABC`, then the mid-point of side BC is

Options

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

Answer

To find the midpoint of side BC of triangle ABC given the vertices A(1, 4, 2), B(-2, 3, -5), and the centroid G(4/3, 0, -2/3), we will first need to determine the coordinates of point C.

### Step-by-Step Solution:

1. **Understanding the Centroid Formula**:
The centroid \( G \) of a triangle with vertices \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
Here, we know \( G \), \( A \), and \( B \), and we need to find \( C \).

2. **Setting Up the Equations**:
Let the coordinates of point C be \( C(x_3, y_3, z_3) \). We can set up the equations based on the centroid formula:
\[ \frac{1 + (-2) + x_3}{3} = \frac{4}{3} \]
\[ \frac{4 + 3 + y_3}{3} = 0 \]
\[ \frac{2 + (-5) + z_3}{3} = -\frac{2}{3} \]

3. **Solving for \( x_3 \)**:
From the first equation:
\[ \frac{-1 + x_3}{3} = \frac{4}{3} \]
Multiply both sides by 3:
\[ -1 + x_3 = 4 \implies x_3 = 5 \]

4. **Solving for \( y_3 \)**:
From the second equation:
\[ \frac{7 + y_3}{3} = 0 \]
Multiply both sides by 3:
\[ 7 + y_3 = 0 \implies y_3 = -7 \]

5. **Solving for \( z_3 \)**:
From the third equation:
\[ \frac{-3 + z_3}{3} = -\frac{2}{3} \]
Multiply both sides by 3:
\[ -3 + z_3 = -2 \implies z_3 = 1 \]

6. **Coordinates of Point C**:
Thus, the coordinates of point C are \( C(5, -7, 1) \).

7. **Finding the Midpoint of BC**:
The midpoint \( D \) of segment BC can be calculated using the midpoint formula:
\[ D = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}, \frac{z_B + z_C}{2} \right) \]
Where \( B(-2, 3, -5) \) and \( C(5, -7, 1) \).

Substituting the values:
\[ D_x = \frac{-2 + 5}{2} = \frac{3}{2} \]
\[ D_y = \frac{3 + (-7)}{2} = \frac{-4}{2} = -2 \]
\[ D_z = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \]

8. **Final Coordinates of Midpoint D**:
Therefore, the coordinates of the midpoint D are:
\[ D\left( \frac{3}{2}, -2, -2 \right) \]

### Conclusion:
The midpoint of side BC is \( D\left( \frac{3}{2}, -2, -2 \right) \).