The mid point of the sides of a triangle are (1,5,-1) (0,4,-2) and (2,3,4) find its vertices also find the centroid of the triangle

Answer

To find the vertices of the triangle given the midpoints of its sides and to find the centroid of the triangle, we can follow these steps:

### Step 1: Set up the midpoints
Let the vertices of the triangle be \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \). The midpoints of the sides of the triangle are given as:
- Midpoint of \( AB \): \( M_1(1, 5, -1) \)
- Midpoint of \( BC \): \( M_2(0, 4, -2) \)
- Midpoint of \( CA \): \( M_3(2, 3, 4) \)

Using the midpoint formula, we can establish the following equations:

1. For midpoint \( M_1 \):
\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = (1, 5, -1) \]
This gives us:
\[ x_1 + x_2 = 2 \quad (1) \]
\[ y_1 + y_2 = 10 \quad (2) \]
\[ z_1 + z_2 = -2 \quad (3) \]

2. For midpoint \( M_2 \):
\[ \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right) = (0, 4, -2) \]
This gives us:
\[ x_2 + x_3 = 0 \quad (4) \]
\[ y_2 + y_3 = 8 \quad (5) \]
\[ z_2 + z_3 = -4 \quad (6) \]

3. For midpoint \( M_3 \):
\[ \left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2}, \frac{z_3 + z_1}{2} \right) = (2, 3, 4) \]
This gives us:
\[ x_3 + x_1 = 4 \quad (7) \]
\[ y_3 + y_1 = 6 \quad (8) \]
\[ z_3 + z_1 = 8 \quad (9) \]

### Step 2: Solve the equations
Now we have a system of equations to solve for \( x_1, x_2, x_3 \), \( y_1, y_2, y_3 \), and \( z_1, z_2, z_3 \).

#### Solving for \( x \):
From equations (1), (4), and (7):
1. From (4): \( x_3 = -x_2 \)
2. Substitute \( x_3 \) into (7):
\[ -x_2 + x_1 = 4 \implies x_1 = x_2 + 4 \quad (10) \]
3. Substitute \( x_1 \) from (10) into (1):
\[ (x_2 + 4) + x_2 = 2 \implies 2x_2 + 4 = 2 \implies 2x_2 = -2 \implies x_2 = -1 \]
4. Now substitute \( x_2 \) back to find \( x_1 \) and \( x_3 \):
\[ x_1 = -1 + 4 = 3 \]
\[ x_3 = -(-1) = 1 \]

#### Solving for \( y \):
From equations (2), (5), and (8):
1. From (5): \( y_3 = 8 - y_2 \)
2. Substitute \( y_3 \) into (8):
\[ (8 - y_2) + y_1 = 6 \implies y_1 - y_2 = -2 \implies y_1 = y_2 - 2 \quad (11) \]
3. Substitute \( y_1 \) from (11) into (2):
\[ (y_2 - 2) + y_2 = 10 \implies 2y_2 - 2 = 10 \implies 2y_2 = 12 \implies y_2 = 6 \]
4. Now substitute \( y_2 \) back to find \( y_1 \) and \( y_3 \):
\[ y_1 = 6 - 2 = 4 \]
\[ y_3 = 8 - 6 = 2 \]

#### Solving for \( z \):
From equations (3), (6), and (9):
1. From (6): \( z_3 = -4 - z_2 \)
2. Substitute \( z_3 \) into (9):
\[ (-4 - z_2) + z_1 = 8 \implies z_1 - z_2 = 12 \implies z_1 = z_2 + 12 \quad (12) \]
3. Substitute \( z_1 \) from (12) into (3):
\[ z_2 + 12 + z_2 = -2 \implies 2z_2 + 12 = -2 \implies 2z_2 = -14 \implies z_2 = -7 \]
4. Now substitute \( z_2 \) back to find \( z_1 \) and \( z_3 \):
\[ z_1 = -7 + 12 = 5 \]
\[ z_3 = -4 - (-7) = 3 \]

### Step 3: Compile the vertices
Now we have the coordinates of the vertices:
- \( A(3, 4, 5) \)
- \( B(-1, 6, -7) \)
- \( C(1, 2, 3) \)

### Step 4: Find the centroid
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:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
Substituting the values:
\[ G\left( \frac{3 - 1 + 1}{3}, \frac{4 + 6 + 2}{3}, \frac{5 - 7 + 3}{3} \right) = G\left( \frac{3}{3}, \frac{12}{3}, \frac{1}{3} \right) = G(1, 4, \frac{1}{3}) \]

### Final Answer:
- The vertices of the triangle are \( A(3, 4, 5) \), \( B(-1, 6, -7) \), and \( C(1, 2, 3) \).
- The centroid of the triangle is \( G(1, 4, \frac{1}{3}) \).