The mid-points of the sides of a triangle are given by (3, 2, -4), (9, 8, -15) and (5, 4, 6). Find
the coordinates of its vertices.

Answer

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

### Step 1: Define the midpoints
Let the midpoints of the sides of the triangle be:
- Midpoint of BC: \( M_{BC} = (3, 2, -4) \)
- Midpoint of AC: \( M_{AC} = (9, 8, -15) \)
- Midpoint of AB: \( M_{AB} = (5, 4, 6) \)

Let the vertices of the triangle be:
- Vertex A: \( A(x_1, y_1, z_1) \)
- Vertex B: \( B(x_2, y_2, z_2) \)
- Vertex C: \( C(x_3, y_3, z_3) \)

### Step 2: Set up equations using midpoints
Using the midpoint formula, we can set up the following equations:

1. For midpoint \( M_{BC} \):
\[ \left( \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}, \frac{x_2 + x_3}{2} \right) = (3, 2, -4) \]
This gives us:
\[ y_2 + y_3 = 6 \quad (1) \]
\[ z_2 + z_3 = -8 \quad (2) \]
\[ x_2 + x_3 = 6 \quad (3) \]

2. For midpoint \( M_{AC} \):
\[ \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}, \frac{z_1 + z_3}{2} \right) = (9, 8, -15) \]
This gives us:
\[ x_1 + x_3 = 18 \quad (4) \]
\[ y_1 + y_3 = 16 \quad (5) \]
\[ z_1 + z_3 = -30 \quad (6) \]

3. For midpoint \( M_{AB} \):
\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = (5, 4, 6) \]
This gives us:
\[ x_1 + x_2 = 10 \quad (7) \]
\[ y_1 + y_2 = 8 \quad (8) \]
\[ z_1 + z_2 = 12 \quad (9) \]

### Step 3: Solve the equations
Now we have a system of equations to solve.

#### Solve for \( x \) coordinates:
From equation (1):
\[ y_2 = 6 - y_3 \quad (10) \]
Substituting \( y_2 \) into equation (8):
\[ y_1 + (6 - y_3) = 8 \implies y_1 - y_3 = 2 \quad (11) \]

From equation (4):
\[ x_3 = 18 - x_1 \quad (12) \]
Substituting \( x_3 \) into equation (3):
\[ x_2 + (18 - x_1) = 6 \implies x_2 - x_1 = -12 \quad (13) \]

From equation (7):
\[ x_2 = 10 - x_1 \quad (14) \]
Substituting \( x_2 \) from (14) into (13):
\[ (10 - x_1) - x_1 = -12 \implies 10 - 2x_1 = -12 \implies 2x_1 = 22 \implies x_1 = 11 \]
Now substituting \( x_1 \) back into (14):
\[ x_2 = 10 - 11 = -1 \]
And substituting \( x_1 \) into (12):
\[ x_3 = 18 - 11 = 7 \]

#### Solve for \( y \) coordinates:
Using \( y_1 \) and \( y_3 \):
From equation (11):
\[ y_1 = 2 + y_3 \]
Substituting \( y_2 \) from (10):
\[ (2 + y_3) + (6 - y_3) = 8 \implies 2 + 6 = 8 \implies y_3 = 10 \]
Then substituting \( y_3 \) back into (10):
\[ y_2 = 6 - 10 = -4 \]
And substituting \( y_3 \) into (11):
\[ y_1 = 2 + 10 = 12 \]

#### Solve for \( z \) coordinates:
Using equations (6) and (9):
From equation (6):
\[ z_3 = -30 - z_1 \]
Substituting \( z_2 \) from (9):
\[ z_1 + z_2 = 12 \implies z_2 = 12 - z_1 \]
Substituting \( z_2 \) into (2):
\[ (12 - z_1) + (-30 - z_1) = -8 \implies -2z_1 - 18 = -8 \implies -2z_1 = 10 \implies z_1 = -5 \]
Now substituting \( z_1 \) back into (9):
\[ z_2 = 12 - (-5) = 17 \]
And substituting \( z_1 \) into (6):
\[ z_3 = -30 - (-5) = -25 \]

### Step 4: Final coordinates
The coordinates of the vertices of the triangle are:
- Vertex A: \( (11, 12, -5) \)
- Vertex B: \( (-1, -4, 17) \)
- Vertex C: \( (7, 10, -25) \)