To find the coordinates of the vertices of a triangle given the midpoints of its sides, we can follow these steps:
### Step 1: Assign Variables
Let the vertices of the triangle be:
- \( A(x_1, y_1) \)
- \( B(x_2, y_2) \)
- \( C(x_3, y_3) \)
The midpoints of the sides are given as:
- Midpoint D of BC: \( (1, 2) \)
- Midpoint E of AC: \( (0, -1) \)
- Midpoint F of AB: \( (2, -1) \)
### Step 2: Set Up Equations for Midpoints
Using the midpoint formula, we can set up the following equations based on the midpoints:
1. For midpoint D:
\[
D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (1, 2)
\]
This gives us two equations:
\[
\frac{x_2 + x_3}{2} = 1 \quad \Rightarrow \quad x_2 + x_3 = 2 \quad \text{(Equation 1)}
\]
\[
\frac{y_2 + y_3}{2} = 2 \quad \Rightarrow \quad y_2 + y_3 = 4 \quad \text{(Equation 2)}
\]
2. For midpoint E:
\[
E = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (0, -1)
\]
This gives us:
\[
\frac{x_1 + x_3}{2} = 0 \quad \Rightarrow \quad x_1 + x_3 = 0 \quad \text{(Equation 3)}
\]
\[
\frac{y_1 + y_3}{2} = -1 \quad \Rightarrow \quad y_1 + y_3 = -2 \quad \text{(Equation 4)}
\]
3. For midpoint F:
\[
F = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (2, -1)
\]
This gives us:
\[
\frac{x_1 + x_2}{2} = 2 \quad \Rightarrow \quad x_1 + x_2 = 4 \quad \text{(Equation 5)}
\]
\[
\frac{y_1 + y_2}{2} = -1 \quad \Rightarrow \quad y_1 + y_2 = -2 \quad \text{(Equation 6)}
\]
### Step 3: Solve the System of Equations
Now we have a system of equations to solve:
From Equations 1, 3, and 5 for \( x \):
1. \( x_2 + x_3 = 2 \) (Equation 1)
2. \( x_1 + x_3 = 0 \) (Equation 3)
3. \( x_1 + x_2 = 4 \) (Equation 5)
We can express \( x_3 \) from Equation 3:
\[
x_3 = -x_1
\]
Substituting into Equation 1:
\[
x_2 - x_1 = 2 \quad \Rightarrow \quad x_2 = x_1 + 2 \quad \text{(Equation 7)}
\]
Now substitute \( x_2 \) from Equation 7 into Equation 5:
\[
x_1 + (x_1 + 2) = 4 \quad \Rightarrow \quad 2x_1 + 2 = 4 \quad \Rightarrow \quad 2x_1 = 2 \quad \Rightarrow \quad x_1 = 1
\]
Now, substituting \( x_1 = 1 \) back into Equation 7:
\[
x_2 = 1 + 2 = 3
\]
And substituting \( x_1 = 1 \) into Equation 3:
\[
x_3 = -1
\]
So we have:
- \( x_1 = 1 \)
- \( x_2 = 3 \)
- \( x_3 = -1 \)
Now, let's solve for \( y \) using Equations 2, 4, and 6:
1. \( y_2 + y_3 = 4 \) (Equation 2)
2. \( y_1 + y_3 = -2 \) (Equation 4)
3. \( y_1 + y_2 = -2 \) (Equation 6)
From Equation 4, express \( y_3 \):
\[
y_3 = -2 - y_1
\]
Substituting into Equation 2:
\[
y_2 + (-2 - y_1) = 4 \quad \Rightarrow \quad y_2 - y_1 = 6 \quad \Rightarrow \quad y_2 = y_1 + 6 \quad \text{(Equation 8)}
\]
Now substitute \( y_2 \) from Equation 8 into Equation 6:
\[
y_1 + (y_1 + 6) = -2 \quad \Rightarrow \quad 2y_1 + 6 = -2 \quad \Rightarrow \quad 2y_1 = -8 \quad \Rightarrow \quad y_1 = -4
\]
Substituting \( y_1 = -4 \) back into Equation 8:
\[
y_2 = -4 + 6 = 2
\]
And substituting \( y_1 = -4 \) into Equation 4:
\[
y_3 = -2 - (-4) = 2
\]
So we have:
- \( y_1 = -4 \)
- \( y_2 = 2 \)
- \( y_3 = 2 \)
### Final Coordinates of the Vertices
Thus, the coordinates of the vertices of the triangle are:
- \( A(1, -4) \)
- \( B(3, 2) \)
- \( C(-1, 2) \)
