The midpoints of the sides BC, CA and AB of a `triangle ABC` are D(2,1), E(-5,7)and F(-5, -5) respectively. Find the equations of the sides of `triangle ABC`.

Options

A. `x-2=0`
B. `6x+7y+65=0`
C. `6x-7y+79=0`
D. All of these

Answer

To find the equations of the sides of triangle ABC given the midpoints D(2,1), E(-5,7), and F(-5,-5), we will first determine the coordinates of the vertices A, B, and C using the midpoint formula. Then, we will derive the equations of the lines AB, BC, and AC.

### Step 1: Identify the midpoints and set up equations

Let the coordinates of vertices A, B, and C be:
- A(x1, y1)
- B(x2, y2)
- C(x3, y3)

The midpoints are given as follows:
- D(2, 1) is the midpoint of AB:
\[ D = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right) \implies \frac{x1 + x2}{2} = 2 \quad \text{and} \quad \frac{y1 + y2}{2} = 1 \]
This gives us:
\[ x1 + x2 = 4 \quad (1) \]
\[ y1 + y2 = 2 \quad (2) \]

- E(-5, 7) is the midpoint of BC:
\[ E = \left( \frac{x2 + x3}{2}, \frac{y2 + y3}{2} \right) \implies \frac{x2 + x3}{2} = -5 \quad \text{and} \quad \frac{y2 + y3}{2} = 7 \]
This gives us:
\[ x2 + x3 = -10 \quad (3) \]
\[ y2 + y3 = 14 \quad (4) \]

- F(-5, -5) is the midpoint of AC:
\[ F = \left( \frac{x1 + x3}{2}, \frac{y1 + y3}{2} \right) \implies \frac{x1 + x3}{2} = -5 \quad \text{and} \quad \frac{y1 + y3}{2} = -5 \]
This gives us:
\[ x1 + x3 = -10 \quad (5) \]
\[ y1 + y3 = -10 \quad (6) \]

### Step 2: Solve the equations

Now we have a system of equations:
1. \( x1 + x2 = 4 \)
2. \( y1 + y2 = 2 \)
3. \( x2 + x3 = -10 \)
4. \( y2 + y3 = 14 \)
5. \( x1 + x3 = -10 \)
6. \( y1 + y3 = -10 \)

#### Solve for x-coordinates:

From equation (1):
\[ x2 = 4 - x1 \quad (7) \]

Substituting (7) into equation (3):
\[ (4 - x1) + x3 = -10 \implies x3 = -10 - 4 + x1 = x1 - 14 \quad (8) \]

Substituting (8) into equation (5):
\[ x1 + (x1 - 14) = -10 \implies 2x1 - 14 = -10 \implies 2x1 = 4 \implies x1 = 2 \]

Now substitute \(x1 = 2\) back into (7):
\[ x2 = 4 - 2 = 2 \]

Substituting \(x1 = 2\) into (8):
\[ x3 = 2 - 14 = -12 \]

So, we have:
- \(x1 = 2\)
- \(x2 = 2\)
- \(x3 = -12\)

#### Solve for y-coordinates:

From equation (2):
\[ y2 = 2 - y1 \quad (9) \]

Substituting (9) into equation (4):
\[ (2 - y1) + y3 = 14 \implies y3 = 14 - 2 + y1 = y1 + 12 \quad (10) \]

Substituting (10) into equation (6):
\[ y1 + (y1 + 12) = -10 \implies 2y1 + 12 = -10 \implies 2y1 = -22 \implies y1 = -11 \]

Now substitute \(y1 = -11\) back into (9):
\[ y2 = 2 - (-11) = 2 + 11 = 13 \]

Substituting \(y1 = -11\) into (10):
\[ y3 = -11 + 12 = 1 \]

So, we have:
- \(y1 = -11\)
- \(y2 = 13\)
- \(y3 = 1\)

### Step 3: Coordinates of vertices

The coordinates of the vertices are:
- A(2, -11)
- B(2, 13)
- C(-12, 1)

### Step 4: Find the equations of the sides

#### Equation of line AB:
Using points A(2, -11) and B(2, 13):
Since both points have the same x-coordinate, the equation of line AB is:
\[ x = 2 \]

#### Equation of line BC:
Using points B(2, 13) and C(-12, 1):
Using the two-point form:
\[ y - 13 = \frac{1 - 13}{-12 - 2}(x - 2) \]
\[ y - 13 = \frac{-12}{-14}(x - 2) \]
\[ y - 13 = \frac{6}{7}(x - 2) \]
\[ y - 13 = \frac{6}{7}x - \frac{12}{7} \]
\[ y = \frac{6}{7}x + \frac{79}{7} \]
Multiplying through by 7:
\[ 7y = 6x + 79 \implies 6x - 7y + 79 = 0 \]

#### Equation of line AC:
Using points A(2, -11) and C(-12, 1):
Using the two-point form:
\[ y + 11 = \frac{1 + 11}{-12 - 2}(x - 2) \]
\[ y + 11 = \frac{12}{-14}(x - 2) \]
\[ y + 11 = -\frac{6}{7}(x - 2) \]
\[ y + 11 = -\frac{6}{7}x + \frac{12}{7} \]
\[ y = -\frac{6}{7}x + \frac{12}{7} - 11 \]
\[ y = -\frac{6}{7}x - \frac{65}{7} \]
Multiplying through by 7:
\[ 7y = -6x - 65 \implies 6x + 7y + 65 = 0 \]

### Final equations of the sides of triangle ABC:
1. Line AB: \(x = 2\)
2. Line BC: \(6x - 7y + 79 = 0\)
3. Line AC: \(6x + 7y + 65 = 0\)