One of the vertices of a triangle whose midpoint of edges are (3,1), (5,6), (-3,2) is :

Options

A. `(-5,-3)`
B. `(1,7)`
C. `(-11,5)`
D. None of these

Answer

To find one of the vertices of a triangle given the midpoints of its edges, we can follow these steps:

### Step 1: Define the Midpoints
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 \( AB = \left(3, 1\right) \)
- Midpoint of \( BC = \left(5, 6\right) \)
- Midpoint of \( AC = \left(-3, 2\right) \)

### Step 2: Set Up the Midpoint Formulas
Using the midpoint formula, we can set up equations for each midpoint:
1. For midpoint \( AB \):
\[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (3, 1) \]
This gives us two equations:
\[ x_1 + x_2 = 6 \quad (1) \]
\[ y_1 + y_2 = 2 \quad (2) \]

2. For midpoint \( BC \):
\[ \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) = (5, 6) \]
This gives us:
\[ x_2 + x_3 = 10 \quad (3) \]
\[ y_2 + y_3 = 12 \quad (4) \]

3. For midpoint \( AC \):
\[ \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = (-3, 2) \]
This gives us:
\[ x_1 + x_3 = -6 \quad (5) \]
\[ y_1 + y_3 = 4 \quad (6) \]

### Step 3: Solve the System of Equations
We have the following equations from above:
1. \( x_1 + x_2 = 6 \) (1)
2. \( x_2 + x_3 = 10 \) (3)
3. \( x_1 + x_3 = -6 \) (5)

#### Solve for \( x_1 \), \( x_2 \), and \( x_3 \)
From equation (1):
\[ x_2 = 6 - x_1 \quad (7) \]

Substituting (7) into equation (3):
\[ (6 - x_1) + x_3 = 10 \]
\[ x_3 = 10 - 6 + x_1 = 4 + x_1 \quad (8) \]

Now substituting (8) into equation (5):
\[ x_1 + (4 + x_1) = -6 \]
\[ 2x_1 + 4 = -6 \]
\[ 2x_1 = -10 \implies x_1 = -5 \]

Now substituting \( x_1 = -5 \) back into (7):
\[ x_2 = 6 - (-5) = 11 \]

And substituting \( x_1 = -5 \) into (8):
\[ x_3 = 4 + (-5) = -1 \]

Thus, we have:
\[ x_1 = -5, \quad x_2 = 11, \quad x_3 = -1 \]

#### Solve for \( y_1 \), \( y_2 \), and \( y_3 \)
Using the same method with equations (2), (4), and (6):
1. \( y_1 + y_2 = 2 \) (2)
2. \( y_2 + y_3 = 12 \) (4)
3. \( y_1 + y_3 = 4 \) (6)

From equation (2):
\[ y_2 = 2 - y_1 \quad (9) \]

Substituting (9) into equation (4):
\[ (2 - y_1) + y_3 = 12 \]
\[ y_3 = 12 - 2 + y_1 = 10 + y_1 \quad (10) \]

Now substituting (10) into equation (6):
\[ y_1 + (10 + y_1) = 4 \]
\[ 2y_1 + 10 = 4 \]
\[ 2y_1 = -6 \implies y_1 = -3 \]

Substituting \( y_1 = -3 \) back into (9):
\[ y_2 = 2 - (-3) = 5 \]

And substituting \( y_1 = -3 \) into (10):
\[ y_3 = 10 + (-3) = 7 \]

Thus, we have:
\[ y_1 = -3, \quad y_2 = 5, \quad y_3 = 7 \]

### Step 4: Final Vertices
The vertices of the triangle are:
- \( A(-5, -3) \)
- \( B(11, 5) \)
- \( C(-1, 7) \)

### Conclusion
One of the vertices of the triangle is \( A(-5, -3) \).

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