The mid-point of the sides of a `DeltaABC` are D(6,1) ,E(3,5) and F(-1,-2) then the coordinates of the vertex opposite to D are

Options

A. (-4,2)
B. (-4,5)
C. (2,5)
D. (10,5)

Answer

To find the coordinates of the vertex opposite to D in triangle ABC where D, E, and F are the midpoints of the sides, we can follow these steps:

### Step 1: Understand the Problem
We have a triangle ABC with midpoints D, E, and F. We need to find the coordinates of the vertex opposite to D. The coordinates of the midpoints are given as:
- D(6, 1)
- E(3, 5)
- F(-1, -2)

### Step 2: Set Up the Coordinates
Let the coordinates of the vertices A, B, and C be:
- A(x1, y1)
- B(x2, y2)
- C(x3, y3)

### Step 3: Use the Midpoint Formula
The midpoint of a line segment joining two points (x1, y1) and (x2, y2) is given by:
\[ \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) \]

### Step 4: Set Up Equations for Midpoints
Using the midpoint formula, we can establish the following equations based on the midpoints:
1. For midpoint D (between A and B):
\[ D\left(6, 1\right) = \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}\right) \]
This gives us:
\[ \frac{x1 + x2}{2} = 6 \quad \text{(1)} \]
\[ \frac{y1 + y2}{2} = 1 \quad \text{(2)} \]

2. For midpoint E (between B and C):
\[ E\left(3, 5\right) = \left(\frac{x2 + x3}{2}, \frac{y2 + y3}{2}\right) \]
This gives us:
\[ \frac{x2 + x3}{2} = 3 \quad \text{(3)} \]
\[ \frac{y2 + y3}{2} = 5 \quad \text{(4)} \]

3. For midpoint F (between A and C):
\[ F\left(-1, -2\right) = \left(\frac{x1 + x3}{2}, \frac{y1 + y3}{2}\right) \]
This gives us:
\[ \frac{x1 + x3}{2} = -1 \quad \text{(5)} \]
\[ \frac{y1 + y3}{2} = -2 \quad \text{(6)} \]

### Step 5: Solve the Equations
Now we will solve these equations step by step.

From equation (1):
\[ x1 + x2 = 12 \quad \text{(7)} \]

From equation (2):
\[ y1 + y2 = 2 \quad \text{(8)} \]

From equation (3):
\[ x2 + x3 = 6 \quad \text{(9)} \]

From equation (4):
\[ y2 + y3 = 10 \quad \text{(10)} \]

From equation (5):
\[ x1 + x3 = -2 \quad \text{(11)} \]

From equation (6):
\[ y1 + y3 = -4 \quad \text{(12)} \]

### Step 6: Substitute and Solve for x-coordinates
From equation (9), we can express \(x3\) in terms of \(x2\):
\[ x3 = 6 - x2 \quad \text{(13)} \]

Substituting equation (13) into equation (11):
\[ x1 + (6 - x2) = -2 \]
\[ x1 - x2 = -8 \quad \text{(14)} \]

Now we have two equations (7) and (14):
1. \(x1 + x2 = 12\) (7)
2. \(x1 - x2 = -8\) (14)

Adding (7) and (14):
\[ 2x1 = 4 \implies x1 = 2 \]

Substituting \(x1 = 2\) into equation (7):
\[ 2 + x2 = 12 \implies x2 = 10 \]

Now substituting \(x2 = 10\) into equation (13):
\[ x3 = 6 - 10 = -4 \]

### Step 7: Solve for y-coordinates
Using equations (8) and (10):
From equation (10):
\[ y3 = 10 - y2 \quad \text{(15)} \]

Substituting equation (15) into equation (12):
\[ y1 + (10 - y2) = -4 \]
\[ y1 - y2 = -14 \quad \text{(16)} \]

Now we have two equations (8) and (16):
1. \(y1 + y2 = 2\) (8)
2. \(y1 - y2 = -14\) (16)

Adding (8) and (16):
\[ 2y1 = -12 \implies y1 = -6 \]

Substituting \(y1 = -6\) into equation (8):
\[ -6 + y2 = 2 \implies y2 = 8 \]

Now substituting \(y2 = 8\) into equation (15):
\[ y3 = 10 - 8 = 2 \]

### Step 8: Final Coordinates
Thus, the coordinates of the vertices A, B, and C are:
- A(2, -6)
- B(10, 8)
- C(-4, 2)

The vertex opposite to D is B, which has coordinates:
\[ \text{Vertex opposite to D} = B(10, 8) \]

### Conclusion
The coordinates of the vertex opposite to D are \( (10, 8) \).