The midpoints of the sides of a triangle are (3, 4) , (4, 1) and (2, 0) . Find the vertices of the triangle .

Answer

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

### Step 1: Assign Variables
Let the vertices of the triangle be A(x1, y1), B(x2, y2), and C(x3, y3). The midpoints of the sides are given as:
- Midpoint D = (3, 4)
- Midpoint E = (4, 1)
- Midpoint F = (2, 0)

### Step 2: Set Up Midpoint Equations
Using the midpoint formula, we can set up equations based on the midpoints:
1. For midpoint D (3, 4):
\[ D = \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) \implies \begin{cases} \frac{x1 + x2}{2} = 3 \implies x1 + x2 = 6 \quad (1) \\ \frac{y1 + y2}{2} = 4 \implies y1 + y2 = 8 \quad (2) \end{cases} \]

2. For midpoint E (4, 1):
\[ E = \left(\frac{x2 + x3}{2}, \frac{y2 + y3}{2}\right) \implies \begin{cases} \frac{x2 + x3}{2} = 4 \implies x2 + x3 = 8 \quad (3) \\ \frac{y2 + y3}{2} = 1 \implies y2 + y3 = 2 \quad (4) \end{cases} \]

3. For midpoint F (2, 0):
\[ F = \left(\frac{x1 + x3}{2}, \frac{y1 + y3}{2}\right) \implies \begin{cases} \frac{x1 + x3}{2} = 2 \implies x1 + x3 = 4 \quad (5) \\ \frac{y1 + y3}{2} = 0 \implies y1 + y3 = 0 \quad (6) \end{cases} \]

### Step 3: Solve the Equations
Now we have a system of equations to solve for x1, y1, x2, y2, x3, and y3.

From equations (1), (3), and (5):
1. From (1): \(x1 + x2 = 6\) (i)
2. From (3): \(x2 + x3 = 8\) (ii)
3. From (5): \(x1 + x3 = 4\) (iii)

We can solve these equations step by step.

**From (i)**:
\[ x2 = 6 - x1 \quad (7) \]

**Substituting (7) into (ii)**:
\[ (6 - x1) + x3 = 8 \implies x3 = 2 + x1 \quad (8) \]

**Substituting (8) into (iii)**:
\[ x1 + (2 + x1) = 4 \implies 2x1 + 2 = 4 \implies 2x1 = 2 \implies x1 = 1 \]

**Finding x2 and x3**:
Substituting \(x1 = 1\) into (7):
\[ x2 = 6 - 1 = 5 \]
Substituting \(x1 = 1\) into (8):
\[ x3 = 2 + 1 = 3 \]

Now we have:
- \(x1 = 1\)
- \(x2 = 5\)
- \(x3 = 3\)

### Step 4: Solve for y-coordinates
Now, let's solve for y-coordinates using equations (2), (4), and (6):
1. From (2): \(y1 + y2 = 8\) (iv)
2. From (4): \(y2 + y3 = 2\) (v)
3. From (6): \(y1 + y3 = 0\) (vi)

**From (iv)**:
\[ y2 = 8 - y1 \quad (9) \]

**Substituting (9) into (v)**:
\[ (8 - y1) + y3 = 2 \implies y3 = 2 - 8 + y1 \implies y3 = y1 - 6 \quad (10) \]

**Substituting (10) into (vi)**:
\[ y1 + (y1 - 6) = 0 \implies 2y1 - 6 = 0 \implies 2y1 = 6 \implies y1 = 3 \]

**Finding y2 and y3**:
Substituting \(y1 = 3\) into (9):
\[ y2 = 8 - 3 = 5 \]
Substituting \(y1 = 3\) into (10):
\[ y3 = 3 - 6 = -3 \]

### Final Coordinates
Now we have:
- \(A(1, 3)\)
- \(B(5, 5)\)
- \(C(3, -3)\)

Thus, the vertices of the triangle are:
- A(1, 3)
- B(5, -1)
- C(3, -3)