To find the vertices A, B, and C of triangle ΔABC given the midpoints of its sides, we can use the midpoint formula. The midpoints of sides BC, CA, and AB are given as follows:
- Midpoint of BC: \( M_{BC} = (3, -3) \)
- Midpoint of CA: \( M_{CA} = (3, -1) \)
- Midpoint of AB: \( M_{AB} = (1, 1) \)
Let the coordinates of the vertices be:
- \( A = (x_1, y_1) \)
- \( B = (x_2, y_2) \)
- \( C = (x_3, y_3) \)
### Step 1: Use the Midpoint Formula
The midpoint formula states that the midpoint \( M \) of a line segment connecting 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 Equations for Midpoints
Using the midpoints provided, we can set up the following equations:
1. For midpoint \( M_{BC} \):
\[
\frac{x_2 + x_3}{2} = 3 \quad \text{and} \quad \frac{y_2 + y_3}{2} = -3
\]
2. For midpoint \( M_{CA} \):
\[
\frac{x_3 + x_1}{2} = 3 \quad \text{and} \quad \frac{y_3 + y_1}{2} = -1
\]
3. For midpoint \( M_{AB} \):
\[
\frac{x_1 + x_2}{2} = 1 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 1
\]
### Step 3: Solve for Coordinates
Now, we can solve these equations step by step.
#### From \( M_{BC} \):
1. \( x_2 + x_3 = 6 \) (Multiplying the first equation by 2)
2. \( y_2 + y_3 = -6 \) (Multiplying the second equation by 2)
#### From \( M_{CA} \):
3. \( x_3 + x_1 = 6 \) (Multiplying the first equation by 2)
4. \( y_3 + y_1 = -2 \) (Multiplying the second equation by 2)
#### From \( M_{AB} \):
5. \( x_1 + x_2 = 2 \) (Multiplying the first equation by 2)
6. \( y_1 + y_2 = 2 \) (Multiplying the second equation by 2)
### Step 4: Substitute and Solve
Now we can substitute equations to find \( x_1, x_2, x_3 \) and \( y_1, y_2, y_3 \).
#### Solving for \( x \)-coordinates:
From equations (1) and (5):
- \( x_2 + x_3 = 6 \) (1)
- \( x_1 + x_2 = 2 \) (5)
Substituting \( x_2 = 2 - x_1 \) into (1):
\[
(2 - x_1) + x_3 = 6 \implies x_3 = 6 - 2 + x_1 = 4 + x_1
\]
Now substitute \( x_3 \) into (3):
\[
(4 + x_1) + x_1 = 6 \implies 2x_1 + 4 = 6 \implies 2x_1 = 2 \implies x_1 = 1
\]
Now substituting \( x_1 = 1 \) back:
- From (5): \( 1 + x_2 = 2 \implies x_2 = 1 \)
- From (1): \( 1 + x_3 = 6 \implies x_3 = 5 \)
Thus, the \( x \)-coordinates are:
- \( x_1 = 1 \)
- \( x_2 = 1 \)
- \( x_3 = 5 \)
#### Solving for \( y \)-coordinates:
Using the same method:
From equations (2) and (6):
- \( y_2 + y_3 = -6 \) (2)
- \( y_1 + y_2 = 2 \) (6)
Substituting \( y_2 = 2 - y_1 \) into (2):
\[
(2 - y_1) + y_3 = -6 \implies y_3 = -6 - 2 + y_1 = -8 + y_1
\]
Now substitute \( y_3 \) into (4):
\[
(-8 + y_1) + y_1 = -2 \implies 2y_1 - 8 = -2 \implies 2y_1 = 6 \implies y_1 = 3
\]
Now substituting \( y_1 = 3 \) back:
- From (6): \( 3 + y_2 = 2 \implies y_2 = -1 \)
- From (2): \( -1 + y_3 = -6 \implies y_3 = -5 \)
Thus, the \( y \)-coordinates are:
- \( y_1 = 3 \)
- \( y_2 = -1 \)
- \( y_3 = -5 \)
### Final Coordinates of Vertices
The vertices of triangle \( \Delta ABC \) are:
- \( A = (1, 3) \)
- \( B = (1, -1) \)
- \( C = (5, -5) \)
### Summary
The vertices of triangle \( \Delta ABC \) are:
- \( A(1, 3) \)
- \( B(1, -1) \)
- \( C(5, -5) \)
