`A(a+1, a-1), B(a^(2)+1, a^(2)-1)` and `C(a^(3)+1,a^(3)-1)` are given points. D (11, 9) is the mid point of AB and E (41, 39) is the mid point of BC. If F is the mid point of AC then `(BF)^(2)` is equal to `K^(2)`, where K =

Answer

To solve the problem, we need to find the value of \( K \) such that \( (BF)^2 = K^2 \). We will follow these steps:

### Step 1: Find the coordinates of points A, B, and C
Given:
- \( A(a+1, a-1) \)
- \( B(a^2+1, a^2-1) \)
- \( C(a^3+1, a^3-1) \)

### Step 2: Use the midpoint formula to find D
The midpoint \( D \) of points \( A \) and \( B \) is given by:
\[ D = \left( \frac{(a+1) + (a^2+1)}{2}, \frac{(a-1) + (a^2-1)}{2} \right) \]
This simplifies to:
\[ D = \left( \frac{a + a^2 + 2}{2}, \frac{a + a^2 - 2}{2} \right) \]

### Step 3: Set the coordinates of D equal to (11, 9)
Since \( D(11, 9) \):
\[ \frac{a + a^2 + 2}{2} = 11 \quad \text{and} \quad \frac{a + a^2 - 2}{2} = 9 \]
From the first equation:
\[ a + a^2 + 2 = 22 \implies a^2 + a - 20 = 0 \]
From the second equation:
\[ a + a^2 - 2 = 18 \implies a^2 + a - 20 = 0 \]
Both equations are the same.

### Step 4: Solve the quadratic equation
The quadratic equation \( a^2 + a - 20 = 0 \) can be factored:
\[ (a - 4)(a + 5) = 0 \]
Thus, \( a = 4 \) or \( a = -5 \).

### Step 5: Find the coordinates of B and C for both values of a
1. For \( a = 4 \):
- \( B(4^2 + 1, 4^2 - 1) = B(17, 15) \)
- \( C(4^3 + 1, 4^3 - 1) = C(65, 63) \)

2. For \( a = -5 \):
- \( B((-5)^2 + 1, (-5)^2 - 1) = B(26, 24) \)
- \( C((-5)^3 + 1, (-5)^3 - 1) = C(-124, -126) \)

### Step 6: Find the midpoint E of BC
Using the midpoint formula for \( B \) and \( C \):
\[ E = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) \]
1. For \( a = 4 \):
\[ E = \left( \frac{17 + 65}{2}, \frac{15 + 63}{2} \right) = \left( 41, 39 \right) \]

2. For \( a = -5 \):
\[ E = \left( \frac{26 - 124}{2}, \frac{24 - 126}{2} \right) = \left( -49, -51 \right) \]

### Step 7: Find the midpoint F of AC
Using the midpoint formula for \( A \) and \( C \):
\[ F = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) \]
1. For \( a = 4 \):
\[ F = \left( \frac{5 + 65}{2}, \frac{3 + 63}{2} \right) = \left( 35, 33 \right) \]

2. For \( a = -5 \):
\[ F = \left( \frac{-4 - 124}{2}, \frac{-6 - 126}{2} \right) = \left( -64, -66 \right) \]

### Step 8: Calculate \( (BF)^2 \)
Using the distance formula:
\[ (BF)^2 = (x_B - x_F)^2 + (y_B - y_F)^2 \]
1. For \( a = 4 \):
\[ (BF)^2 = (17 - 35)^2 + (15 - 33)^2 = (-18)^2 + (-18)^2 = 324 + 324 = 648 \]

2. For \( a = -5 \):
\[ (BF)^2 = (26 + 64)^2 + (24 + 66)^2 = (90)^2 + (90)^2 = 8100 + 8100 = 16200 \]

### Step 9: Find \( K \)
From the calculations:
1. For \( a = 4 \):
\[ K^2 = 648 \implies K = \sqrt{648} = 18\sqrt{2} \]

2. For \( a = -5 \):
\[ K^2 = 16200 \implies K = \sqrt{16200} = 90\sqrt{2} \]

### Final Answer
The value of \( K \) for \( a = 4 \) is \( 18\sqrt{2} \) and for \( a = -5 \) is \( 90\sqrt{2} \).