To find the vertices of the triangle given the midpoints of its sides, we will use the midpoint formula. The midpoints of the sides of the triangle are given as follows:
- Midpoint P (AB) = (1, 5, -1)
- Midpoint Q (BC) = (0, 4, -2)
- Midpoint R (AC) = (2, 3, 4)
Let the vertices of the triangle be A (x1, y1, z1), B (x2, y2, z2), and C (x3, y3, z3).
### Step 1: Set up the equations using the midpoint formula
The midpoint formula states that the coordinates of the midpoint of a line segment connecting points (x1, y1, z1) and (x2, y2, z2) is given by:
\[
\text{Midpoint} = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2} \right)
\]
Using this formula, we can set up the following equations for the midpoints:
1. For midpoint P (AB):
\[
P = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2} \right) = (1, 5, -1)
\]
This gives us:
\[
x1 + x2 = 2 \cdot 1 = 2 \quad \text{(Equation 1)}
\]
\[
y1 + y2 = 2 \cdot 5 = 10 \quad \text{(Equation 2)}
\]
\[
z1 + z2 = 2 \cdot (-1) = -2 \quad \text{(Equation 3)}
\]
2. For midpoint Q (BC):
\[
Q = \left( \frac{x2 + x3}{2}, \frac{y2 + y3}{2}, \frac{z2 + z3}{2} \right) = (0, 4, -2)
\]
This gives us:
\[
x2 + x3 = 2 \cdot 0 = 0 \quad \text{(Equation 4)}
\]
\[
y2 + y3 = 2 \cdot 4 = 8 \quad \text{(Equation 5)}
\]
\[
z2 + z3 = 2 \cdot (-2) = -4 \quad \text{(Equation 6)}
\]
3. For midpoint R (AC):
\[
R = \left( \frac{x1 + x3}{2}, \frac{y1 + y3}{2}, \frac{z1 + z3}{2} \right) = (2, 3, 4)
\]
This gives us:
\[
x1 + x3 = 2 \cdot 2 = 4 \quad \text{(Equation 7)}
\]
\[
y1 + y3 = 2 \cdot 3 = 6 \quad \text{(Equation 8)}
\]
\[
z1 + z3 = 2 \cdot 4 = 8 \quad \text{(Equation 9)}
\]
### Step 2: Solve the equations
Now we have a system of equations to solve for the vertices A, B, and C.
#### Solving for x-coordinates:
From Equations 1, 4, and 7:
1. \( x1 + x2 = 2 \) (Equation 1)
2. \( x2 + x3 = 0 \) (Equation 4)
3. \( x1 + x3 = 4 \) (Equation 7)
From Equation 4, we can express \( x2 \):
\[
x2 = -x3 \quad \text{(Equation 10)}
\]
Substituting Equation 10 into Equation 1:
\[
x1 - x3 = 2 \quad \text{(Equation 11)}
\]
Now substituting Equation 10 into Equation 7:
\[
x1 - x3 = 4 \quad \text{(Equation 12)}
\]
Now we have:
1. \( x1 - x3 = 2 \) (Equation 11)
2. \( x1 - x3 = 4 \) (Equation 12)
This is incorrect; let's solve it correctly.
From Equation 1:
\[
x1 + x2 = 2 \quad \text{(1)}
\]
From Equation 4:
\[
x2 + x3 = 0 \quad \text{(4)}
\]
From Equation 7:
\[
x1 + x3 = 4 \quad \text{(7)}
\]
Substituting \( x2 = 2 - x1 \) into Equation 4:
\[
(2 - x1) + x3 = 0 \implies x3 = x1 - 2 \quad \text{(Equation 13)}
\]
Now substituting Equation 13 into Equation 7:
\[
x1 + (x1 - 2) = 4 \implies 2x1 - 2 = 4 \implies 2x1 = 6 \implies x1 = 3
\]
Now substituting \( x1 = 3 \) into Equation 1:
\[
3 + x2 = 2 \implies x2 = -1
\]
And substituting \( x1 = 3 \) into Equation 13:
\[
x3 = 3 - 2 = 1
\]
Thus, we have:
- \( x1 = 3 \)
- \( x2 = -1 \)
- \( x3 = 1 \)
#### Solving for y-coordinates:
Using the same process for y-coordinates:
1. \( y1 + y2 = 10 \) (Equation 2)
2. \( y2 + y3 = 8 \) (Equation 5)
3. \( y1 + y3 = 6 \) (Equation 8)
From Equation 5:
\[
y2 = 8 - y3 \quad \text{(Equation 14)}
\]
Substituting Equation 14 into Equation 2:
\[
y1 + (8 - y3) = 10 \implies y1 - y3 = 2 \quad \text{(Equation 15)}
\]
Substituting Equation 14 into Equation 8:
\[
y1 + y3 = 6 \quad \text{(Equation 16)}
\]
Now we have:
1. \( y1 - y3 = 2 \) (Equation 15)
2. \( y1 + y3 = 6 \) (Equation 16)
Adding these two equations:
\[
2y1 = 8 \implies y1 = 4
\]
Substituting \( y1 = 4 \) into Equation 15:
\[
4 - y3 = 2 \implies y3 = 2
\]
Substituting \( y1 = 4 \) into Equation 14:
\[
y2 = 8 - 2 = 6
\]
Thus, we have:
- \( y1 = 4 \)
- \( y2 = 6 \)
- \( y3 = 2 \)
#### Solving for z-coordinates:
Using the same process for z-coordinates:
1. \( z1 + z2 = -2 \) (Equation 3)
2. \( z2 + z3 = -4 \) (Equation 6)
3. \( z1 + z3 = 8 \) (Equation 9)
From Equation 6:
\[
z2 = -4 - z3 \quad \text{(Equation 17)}
\]
Substituting Equation 17 into Equation 3:
\[
z1 + (-4 - z3) = -2 \implies z1 - z3 = 2 \quad \text{(Equation 18)}
\]
Substituting Equation 17 into Equation 9:
\[
z1 + z3 = 8 \quad \text{(Equation 19)}
\]
Now we have:
1. \( z1 - z3 = 2 \) (Equation 18)
2. \( z1 + z3 = 8 \) (Equation 19)
Adding these two equations:
\[
2z1 = 10 \implies z1 = 5
\]
Substituting \( z1 = 5 \) into Equation 18:
\[
5 - z3 = 2 \implies z3 = 3
\]
Substituting \( z1 = 5 \) into Equation 17:
\[
z2 = -4 - 3 = -7
\]
Thus, we have:
- \( z1 = 5 \)
- \( z2 = -7 \)
- \( z3 = 3 \)
### Final vertices of the triangle
The vertices of the triangle are:
- A (3, 4, 5)
- B (-1, 6, -7)
- C (1, 2, 3)
### Summary of the solution steps:
1. Set up equations using the midpoint formula for each side of the triangle.
2. Solve the equations systematically for each coordinate (x, y, z).
3. Combine results to find the final coordinates of the vertices.
