The value of k for which points (k, 1), (2, 3), (3, 4) are collinear.

Answer

To find the value of \( k \) for which the points \( (k, 1) \), \( (2, 3) \), and \( (3, 4) \) are collinear, we can use the concept of determinants. The points are collinear if the area of the triangle formed by them is zero, which can be determined using the determinant of a matrix formed by their coordinates.

### Step-by-Step Solution:

1. **Set up the determinant**: The points can be represented in the following matrix form:
\[ \begin{vmatrix} k & 1 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 1 \end{vmatrix} \]
The determinant of this matrix should equal zero for the points to be collinear.

2. **Calculate the determinant**:
Using the determinant formula for a 3x3 matrix, we have:
\[ \text{Det} = k \begin{vmatrix} 3 & 1 \\ 4 & 1 \end{vmatrix} - 1 \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} \]

Now, we calculate each of the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} 3 & 1 \\ 4 & 1 \end{vmatrix} = (3 \cdot 1) - (4 \cdot 1) = 3 - 4 = -1 \]
- For the second determinant:
\[ \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = (2 \cdot 1) - (3 \cdot 1) = 2 - 3 = -1 \]
- For the third determinant:
\[ \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} = (2 \cdot 4) - (3 \cdot 3) = 8 - 9 = -1 \]

3. **Substituting back into the determinant**:
Now substituting these values back into the determinant:
\[ \text{Det} = k(-1) - 1(-1) + 1(-1) = -k + 1 - 1 = -k \]

4. **Setting the determinant to zero**:
For the points to be collinear, we set the determinant equal to zero:
\[ -k = 0 \]

5. **Solving for \( k \)**:
Thus, we find:
\[ k = 0 \]

### Conclusion:
The value of \( k \) for which the points \( (k, 1) \), \( (2, 3) \), and \( (3, 4) \) are collinear is \( k = 0 \).