If coordinates of the mid-points of the sides of a triangle are `(1, -1)`, (2, 3) and (3, 2), then area of the trangle is

Options

A. 5 sq. units
B. 6 sq. units
C. `(11)/(2)` sq. units
D. None of these

Answer

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

### Step 1: Identify the Midpoints
Let the midpoints of the sides of the triangle be:
- D(1, -1)
- E(2, 3)
- F(3, 2)

### Step 2: Use the Midpoint Formula
The coordinates of the midpoints can be expressed in terms of the coordinates of the vertices of the triangle. Let the vertices be A(x1, y1), B(x2, y2), and C(x3, y3). The midpoint coordinates can be expressed as:
- D = ((x1 + x2)/2, (y1 + y2)/2) = (1, -1)
- E = ((x2 + x3)/2, (y2 + y3)/2) = (2, 3)
- F = ((x1 + x3)/2, (y1 + y3)/2) = (3, 2)

### Step 3: Set Up the Equations
From the midpoint coordinates, we can set up the following equations:
1. \( \frac{x1 + x2}{2} = 1 \) → \( x1 + x2 = 2 \) (Equation 1)
2. \( \frac{y1 + y2}{2} = -1 \) → \( y1 + y2 = -2 \) (Equation 2)
3. \( \frac{x2 + x3}{2} = 2 \) → \( x2 + x3 = 4 \) (Equation 3)
4. \( \frac{y2 + y3}{2} = 3 \) → \( y2 + y3 = 6 \) (Equation 4)
5. \( \frac{x1 + x3}{2} = 3 \) → \( x1 + x3 = 6 \) (Equation 5)
6. \( \frac{y1 + y3}{2} = 2 \) → \( y1 + y3 = 4 \) (Equation 6)

### Step 4: Solve the System of Equations
Now, we can solve these equations step by step.

From Equation 1:
- \( x2 = 2 - x1 \) (Substituting into Equation 3)

Substituting \( x2 \) into Equation 3:
- \( (2 - x1) + x3 = 4 \)
- \( x3 = 4 - 2 + x1 \)
- \( x3 = 2 + x1 \) (Equation 7)

Now substitute \( x3 \) from Equation 7 into Equation 5:
- \( x1 + (2 + x1) = 6 \)
- \( 2x1 + 2 = 6 \)
- \( 2x1 = 4 \)
- \( x1 = 2 \)

Now substitute \( x1 = 2 \) back into Equation 1 and Equation 7:
- From Equation 1: \( x2 = 2 - 2 = 0 \)
- From Equation 7: \( x3 = 2 + 2 = 4 \)

Now we have:
- \( x1 = 2, x2 = 0, x3 = 4 \)

Next, we solve for \( y1, y2, y3 \) using Equations 2, 4, and 6.

From Equation 2:
- \( y2 = -2 - y1 \) (Substituting into Equation 4)

Substituting \( y2 \) into Equation 4:
- \( (-2 - y1) + y3 = 6 \)
- \( y3 = 6 + 2 + y1 \)
- \( y3 = 8 + y1 \) (Equation 8)

Now substitute \( y3 \) from Equation 8 into Equation 6:
- \( y1 + (8 + y1) = 4 \)
- \( 2y1 + 8 = 4 \)
- \( 2y1 = -4 \)
- \( y1 = -2 \)

Now substitute \( y1 = -2 \) back into Equation 2 and Equation 8:
- From Equation 2: \( y2 = -2 - (-2) = 0 \)
- From Equation 8: \( y3 = 8 + (-2) = 6 \)

Now we have:
- \( y1 = -2, y2 = 0, y3 = 6 \)

### Step 5: Coordinates of the Vertices
The coordinates of the vertices of the triangle are:
- A(2, -2)
- B(0, 0)
- C(4, 6)

### Step 6: Calculate the Area of the Triangle
Using the formula for the area of a triangle given vertices \((x1, y1)\), \((x2, y2)\), \((x3, y3)\):
\[ \text{Area} = \frac{1}{2} \left| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) \right| \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \left| 2(0 - 6) + 0(6 + 2) + 4(-2 - 0) \right| \]
\[ = \frac{1}{2} \left| 2(-6) + 0 + 4(-2) \right| \]
\[ = \frac{1}{2} \left| -12 - 8 \right| \]
\[ = \frac{1}{2} \left| -20 \right| = \frac{20}{2} = 10 \]

### Final Answer
The area of the triangle is \(10\) square units.

---