The points `(x +1, 2), (1, x +2), ((1)/(x+1),(2)/(x+1))` are collinear, then x is equal to

Options

A. `-4`
B. `-8`
C. 4
D. 8

Answer

To determine the value of \( x \) for which the points \( (x + 1, 2) \), \( (1, x + 2) \), and \( \left(\frac{1}{x + 1}, \frac{2}{x + 1}\right) \) are collinear, we can use the concept of the area of a triangle formed by three points. If the area is zero, the points are collinear.

### Step-by-Step Solution:

1. **Set up the points**:
Let the points be:
- \( A = (x + 1, 2) \)
- \( B = (1, x + 2) \)
- \( C = \left(\frac{1}{x + 1}, \frac{2}{x + 1}\right) \)

2. **Use the determinant method**:
The points \( A \), \( B \), and \( C \) are collinear if the following determinant is equal to zero:
\[ \begin{vmatrix} x + 1 & 2 & 1 \\ 1 & x + 2 & 1 \\ \frac{1}{x + 1} & \frac{2}{x + 1} & 1 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expand the determinant:
\[ = (x + 1) \begin{vmatrix} x + 2 & 1 \\ \frac{2}{x + 1} & 1 \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ \frac{1}{x + 1} & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & x + 2 \\ \frac{1}{x + 1} & \frac{2}{x + 1} \end{vmatrix} \]

4. **Calculate the 2x2 determinants**:
- First determinant:
\[ \begin{vmatrix} x + 2 & 1 \\ \frac{2}{x + 1} & 1 \end{vmatrix} = (x + 2) \cdot 1 - 1 \cdot \frac{2}{x + 1} = x + 2 - \frac{2}{x + 1} \]
- Second determinant:
\[ \begin{vmatrix} 1 & 1 \\ \frac{1}{x + 1} & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot \frac{1}{x + 1} = 1 - \frac{1}{x + 1} \]
- Third determinant:
\[ \begin{vmatrix} 1 & x + 2 \\ \frac{1}{x + 1} & \frac{2}{x + 1} \end{vmatrix} = 1 \cdot \frac{2}{x + 1} - (x + 2) \cdot \frac{1}{x + 1} = \frac{2 - (x + 2)}{x + 1} = \frac{0 - x}{x + 1} = \frac{-x}{x + 1} \]

5. **Substituting back into the determinant**:
Now substituting back into the determinant:
\[ (x + 1) \left(x + 2 - \frac{2}{x + 1}\right) - 2 \left(1 - \frac{1}{x + 1}\right) + \frac{-x}{x + 1} = 0 \]

6. **Simplifying the equation**:
Multiply through by \( x + 1 \) to eliminate the fraction:
\[ (x + 1)(x + 2)(x + 1) - 2(x + 1)(1 - \frac{1}{x + 1}) - x = 0 \]
Simplifying gives:
\[ (x + 1)(x^2 + 3x + 2) - 2(x) = 0 \]
Expanding and simplifying leads to:
\[ x^3 + 4x^2 - 2x = 0 \]

7. **Factoring the polynomial**:
Factor out \( x \):
\[ x(x^2 + 4x - 2) = 0 \]
This gives \( x = 0 \) or solving \( x^2 + 4x - 2 = 0 \) using the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{16 + 8}}{2} = \frac{-4 \pm \sqrt{24}}{2} = \frac{-4 \pm 2\sqrt{6}}{2} = -2 \pm \sqrt{6} \]

### Final Answer:
Thus, the values of \( x \) for which the points are collinear are:
\[ x = 0, \quad x = -2 + \sqrt{6}, \quad x = -2 - \sqrt{6} \]