The midpoints of the sides BC, CA and AB of a `Delta ABC` are D(3, 4), E(8, 9) and F(6, 7) respectively. Find the coordinates of the vertices of the triangle.

Answer

To find the coordinates of the vertices of triangle ABC given the midpoints D, E, and F of sides BC, CA, and AB respectively, we can use the midpoint formula.

### Step-by-Step Solution:

1. **Identify the Midpoints**:
- Let the coordinates of points A, B, and C be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \).
- The midpoints are given as:
- D(3, 4) is the midpoint of BC.
- E(8, 9) is the midpoint of CA.
- F(6, 7) is the midpoint of AB.

2. **Use the Midpoint Formula**:
- The midpoint formula states that the midpoint \( M \) of a line segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

3. **Set Up Equations for Each Midpoint**:
- For midpoint D(3, 4):
\[ D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (3, 4) \]
This gives us two equations:
\[ x_2 + x_3 = 6 \quad (1) \]
\[ y_2 + y_3 = 8 \quad (2) \]

- For midpoint E(8, 9):
\[ E = \left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2} \right) = (8, 9) \]
This gives us:
\[ x_3 + x_1 = 16 \quad (3) \]
\[ y_3 + y_1 = 18 \quad (4) \]

- For midpoint F(6, 7):
\[ F = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (6, 7) \]
This gives us:
\[ x_1 + x_2 = 12 \quad (5) \]
\[ y_1 + y_2 = 14 \quad (6) \]

4. **Solve the System of Equations**:
- We have the following system of equations:
- From (1) and (5):
\[ x_2 + x_3 = 6 \quad (1) \]
\[ x_1 + x_2 = 12 \quad (5) \]
Substituting \( x_3 = 6 - x_2 \) into (3):
\[ (6 - x_2) + x_1 = 16 \implies x_1 - x_2 = 10 \quad (7) \]
Now we have:
\[ x_1 + x_2 = 12 \quad (5) \]
\[ x_1 - x_2 = 10 \quad (7) \]
Adding (5) and (7):
\[ 2x_1 = 22 \implies x_1 = 11 \]
Substituting \( x_1 = 11 \) into (5):
\[ 11 + x_2 = 12 \implies x_2 = 1 \]
Now substituting \( x_2 = 1 \) into (1):
\[ 1 + x_3 = 6 \implies x_3 = 5 \]

- Now for the y-coordinates:
- From (2) and (6):
\[ y_2 + y_3 = 8 \quad (2) \]
\[ y_1 + y_2 = 14 \quad (6) \]
Substituting \( y_3 = 8 - y_2 \) into (4):
\[ (8 - y_2) + y_1 = 18 \implies y_1 - y_2 = 10 \quad (8) \]
Now we have:
\[ y_1 + y_2 = 14 \quad (6) \]
\[ y_1 - y_2 = 10 \quad (8) \]
Adding (6) and (8):
\[ 2y_1 = 24 \implies y_1 = 12 \]
Substituting \( y_1 = 12 \) into (6):
\[ 12 + y_2 = 14 \implies y_2 = 2 \]
Now substituting \( y_2 = 2 \) into (2):
\[ 2 + y_3 = 8 \implies y_3 = 6 \]

5. **Final Coordinates**:
- The coordinates of the vertices A, B, and C are:
- \( A(11, 12) \)
- \( B(1, 2) \)
- \( C(5, 6) \)

### Summary of Coordinates:
- A = (11, 12)
- B = (1, 2)
- C = (5, 6)