The mid-points of the sides of a triangle ABC are given by A(0, 0, 0), B(2, -1, 3) and
C(4, 5, 8) Find the coordinates of A, B and C.

Answer

To find the coordinates of the vertices A, B, and C of triangle ABC given the midpoints of its sides, we can use the midpoint formula. The midpoints are given as follows:

- Midpoint A (P) = (0, 0, 0)
- Midpoint B (Q) = (2, -1, 3)
- Midpoint C (R) = (4, 5, 8)

Let the coordinates of the vertices be:
- A = (x1, y1, z1)
- B = (x2, y2, z2)
- C = (x3, y3, z3)

### Step 1: Set up equations using the midpoint formula

1. For midpoint A (P) which is the midpoint of AB:
\[ P = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2} \right) = (0, 0, 0) \]
This gives us three equations:
\[ \frac{x1 + x2}{2} = 0 \implies x1 + x2 = 0 \quad \text{(1)} \]
\[ \frac{y1 + y2}{2} = 0 \implies y1 + y2 = 0 \quad \text{(2)} \]
\[ \frac{z1 + z2}{2} = 0 \implies z1 + z2 = 0 \quad \text{(3)} \]

2. For midpoint B (Q) which is the midpoint of BC:
\[ Q = \left( \frac{x2 + x3}{2}, \frac{y2 + y3}{2}, \frac{z2 + z3}{2} \right) = (2, -1, 3) \]
This gives us:
\[ \frac{x2 + x3}{2} = 2 \implies x2 + x3 = 4 \quad \text{(4)} \]
\[ \frac{y2 + y3}{2} = -1 \implies y2 + y3 = -2 \quad \text{(5)} \]
\[ \frac{z2 + z3}{2} = 3 \implies z2 + z3 = 6 \quad \text{(6)} \]

3. For midpoint C (R) which is the midpoint of AC:
\[ R = \left( \frac{x3 + x1}{2}, \frac{y3 + y1}{2}, \frac{z3 + z1}{2} \right) = (4, 5, 8) \]
This gives us:
\[ \frac{x3 + x1}{2} = 4 \implies x3 + x1 = 8 \quad \text{(7)} \]
\[ \frac{y3 + y1}{2} = 5 \implies y3 + y1 = 10 \quad \text{(8)} \]
\[ \frac{z3 + z1}{2} = 8 \implies z3 + z1 = 16 \quad \text{(9)} \]

### Step 2: Solve the equations

Now we have a system of equations:

From equations (1), (4), and (7):
1. \( x1 + x2 = 0 \) (1)
2. \( x2 + x3 = 4 \) (4)
3. \( x3 + x1 = 8 \) (7)

Substituting \( x2 = -x1 \) from (1) into (4):
\[ -x1 + x3 = 4 \implies x3 = x1 + 4 \quad \text{(10)} \]
Now substituting (10) into (7):
\[ (x1 + 4) + x1 = 8 \implies 2x1 + 4 = 8 \implies 2x1 = 4 \implies x1 = 2 \]
Now substituting \( x1 = 2 \) back into (1) and (10):
\[ x2 = -2, \quad x3 = 6 \]

Now we have \( x1 = 2, x2 = -2, x3 = 6 \).

Next, we can apply the same process for y-coordinates using equations (2), (5), and (8):
1. \( y1 + y2 = 0 \) (2)
2. \( y2 + y3 = -2 \) (5)
3. \( y3 + y1 = 10 \) (8)

Substituting \( y2 = -y1 \) into (5):
\[ -y1 + y3 = -2 \implies y3 = y1 - 2 \quad \text{(11)} \]
Substituting (11) into (8):
\[ (y1 - 2) + y1 = 10 \implies 2y1 - 2 = 10 \implies 2y1 = 12 \implies y1 = 6 \]
Now substituting \( y1 = 6 \) back into (2) and (11):
\[ y2 = -6, \quad y3 = 4 \]

Now we have \( y1 = 6, y2 = -6, y3 = 4 \).

Finally, we solve for z-coordinates using equations (3), (6), and (9):
1. \( z1 + z2 = 0 \) (3)
2. \( z2 + z3 = 6 \) (6)
3. \( z3 + z1 = 16 \) (9)

Substituting \( z2 = -z1 \) into (6):
\[ -z1 + z3 = 6 \implies z3 = z1 + 6 \quad \text{(12)} \]
Substituting (12) into (9):
\[ (z1 + 6) + z1 = 16 \implies 2z1 + 6 = 16 \implies 2z1 = 10 \implies z1 = 5 \]
Now substituting \( z1 = 5 \) back into (3) and (12):
\[ z2 = -5, \quad z3 = 11 \]

### Final Coordinates

Thus, the coordinates of the vertices A, B, and C are:
- A = (2, 6, 5)
- B = (-2, -6, -5)
- C = (6, 4, 11)