If the points A(2, 3), B(4, k) and C(6, -3) are collinear, find the value of k.

Explanation

To find the value of \( k \) such that the points \( A(2, 3) \), \( B(4, k) \), and \( C(6, -3) \) are collinear, we can use the concept of the area of the triangle formed by these three points. If the area is zero, then the points are collinear.

### Step-by-Step Solution:

1. **Identify the points**:
- Let \( A(2, 3) \) be \( (x_1, y_1) \)
- Let \( B(4, k) \) be \( (x_2, y_2) \)
- Let \( C(6, -3) \) be \( (x_3, y_3) \)

2. **Use the formula for the area of a triangle**:
The area \( \Delta \) formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Since we want the points to be collinear, we set the area to zero:
\[ \Delta = 0 \]

3. **Substitute the coordinates into the area formula**:
Substituting the coordinates of points \( A \), \( B \), and \( C \):
\[ 0 = \frac{1}{2} \left| 2(k - (-3)) + 4(-3 - 3) + 6(3 - k) \right| \]
Simplifying this:
\[ 0 = \frac{1}{2} \left| 2(k + 3) + 4(-6) + 6(3 - k) \right| \]
\[ 0 = \frac{1}{2} \left| 2k + 6 - 24 + 18 - 6k \right| \]
\[ 0 = \frac{1}{2} \left| -4k + 0 \right| \]
\[ 0 = \left| -4k \right| \]

4. **Solve for \( k \)**:
Since the absolute value is zero, we have:
\[ -4k = 0 \implies k = 0 \]

### Conclusion:
The value of \( k \) is \( 0 \).