The line `(K+1)^(2)x+Ky-2K^(2)-2=0` passes through the point (m, n) for all real values of K, then

Options

A. `m+n=2`
B. `m-n=6`
C. `(m)/(n)=2`
D. `(m)/(n)=(1)/(2)`

Answer

To solve the problem, we need to analyze the given line equation and find the conditions under which it passes through the point (m, n) for all real values of K.

### Step-by-step Solution:

1. **Write the given equation**:
The equation of the line is given as:
\[ (K+1)^2 x + K y - 2K^2 - 2 = 0 \]

2. **Expand the equation**:
We can expand \((K+1)^2\) using the formula \(a^2 + 2ab + b^2\):
\[ (K^2 + 2K + 1)x + Ky - 2K^2 - 2 = 0 \]
This simplifies to:
\[ K^2 x + 2K x + x + K y - 2K^2 - 2 = 0 \]

3. **Group the terms**:
Rearranging the equation, we can group the terms involving K:
\[ K^2(x - 2) + K(2x + y) + (x - 2) = 0 \]

4. **Factor out common terms**:
We can factor out \((x - 2)\) from the equation:
\[ (x - 2)(K^2 + 1) + K(2x + y) = 0 \]

5. **Analyze the equation**:
For the line to pass through the point (m, n) for all values of K, the coefficients of K must equal zero. This gives us two conditions:
- \(x - 2 = 0\) (which gives \(x = 2\))
- \(2x + y = 0\)

6. **Substituting \(x = 2\)**:
Substitute \(x = 2\) into the second condition:
\[ 2(2) + y = 0 \implies 4 + y = 0 \implies y = -4 \]

7. **Conclusion**:
Therefore, the point (m, n) that the line passes through for all values of K is:
\[ (m, n) = (2, -4) \]

### Final Answer:
The point (m, n) is \((2, -4)\).