To find one of the vertices of a triangle given the midpoints of its edges, we can follow these steps:
### Step 1: Understand the Midpoint Formula
The midpoint \( M \) of a line segment connecting two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
### Step 2: Set Up the Problem
Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \). The midpoints of the edges are given as:
- Midpoint of \( AB \) is \( D(3, 1) \)
- Midpoint of \( BC \) is \( E(5, 6) \)
- Midpoint of \( CA \) is \( F(-3, 2) \)
### Step 3: Write Equations for Midpoints
Using the midpoint formula, we can write the following equations:
1. For midpoint \( D \):
\[
\frac{x_1 + x_2}{2} = 3 \quad \Rightarrow \quad x_1 + x_2 = 6 \quad \text{(Equation 1)}
\]
\[
\frac{y_1 + y_2}{2} = 1 \quad \Rightarrow \quad y_1 + y_2 = 2 \quad \text{(Equation 2)}
\]
2. For midpoint \( E \):
\[
\frac{x_2 + x_3}{2} = 5 \quad \Rightarrow \quad x_2 + x_3 = 10 \quad \text{(Equation 3)}
\]
\[
\frac{y_2 + y_3}{2} = 6 \quad \Rightarrow \quad y_2 + y_3 = 12 \quad \text{(Equation 4)}
\]
3. For midpoint \( F \):
\[
\frac{x_1 + x_3}{2} = -3 \quad \Rightarrow \quad x_1 + x_3 = -6 \quad \text{(Equation 5)}
\]
\[
\frac{y_1 + y_3}{2} = 2 \quad \Rightarrow \quad y_1 + y_3 = 4 \quad \text{(Equation 6)}
\]
### Step 4: Solve the Equations
Now we have a system of equations to solve for \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \).
#### For x-coordinates:
1. From Equations 1, 3, and 5:
- Adding Equations 1, 3, and 5:
\[
(x_1 + x_2) + (x_2 + x_3) + (x_1 + x_3) = 6 + 10 - 6
\]
\[
2x_1 + 2x_2 + 2x_3 = 10 \quad \Rightarrow \quad x_1 + x_2 + x_3 = 5 \quad \text{(Equation 7)}
\]
2. Now substitute \( x_2 \) from Equation 1 into Equation 3:
\[
6 - x_1 + x_3 = 10 \quad \Rightarrow \quad x_3 = 4 + x_1
\]
3. Substitute \( x_3 \) into Equation 5:
\[
x_1 + (4 + x_1) = -6 \quad \Rightarrow \quad 2x_1 + 4 = -6 \quad \Rightarrow \quad 2x_1 = -10 \quad \Rightarrow \quad x_1 = -5
\]
4. Now substitute \( x_1 = -5 \) back into Equation 1:
\[
-5 + x_2 = 6 \quad \Rightarrow \quad x_2 = 11
\]
5. Substitute \( x_2 = 11 \) into Equation 3:
\[
11 + x_3 = 10 \quad \Rightarrow \quad x_3 = -1
\]
#### For y-coordinates:
1. Using the same approach with Equations 2, 4, and 6:
- Adding Equations 2, 4, and 6:
\[
(y_1 + y_2) + (y_2 + y_3) + (y_1 + y_3) = 2 + 12 + 4
\]
\[
2y_1 + 2y_2 + 2y_3 = 18 \quad \Rightarrow \quad y_1 + y_2 + y_3 = 9 \quad \text{(Equation 8)}
\]
2. Substitute \( y_2 \) from Equation 2 into Equation 4:
\[
2 - y_1 + y_3 = 12 \quad \Rightarrow \quad y_3 = 10 + y_1
\]
3. Substitute \( y_3 \) into Equation 6:
\[
y_1 + (10 + y_1) = 4 \quad \Rightarrow \quad 2y_1 + 10 = 4 \quad \Rightarrow \quad 2y_1 = -6 \quad \Rightarrow \quad y_1 = -3
\]
4. Substitute \( y_1 = -3 \) back into Equation 2:
\[
-3 + y_2 = 2 \quad \Rightarrow \quad y_2 = 5
\]
5. Substitute \( y_2 = 5 \) into Equation 4:
\[
5 + y_3 = 12 \quad \Rightarrow \quad y_3 = 7
\]
### Final Coordinates
Thus, the coordinates of the vertices are:
- \( A(-5, -3) \)
- \( B(11, 5) \)
- \( C(-1, 7) \)
### Conclusion
One of the vertices of the triangle is \( A(-5, -3) \).
---
