If the three points (k,2k ) (2k,3k) (3,1 ) are collinear, then k is equal to

Options

A. `-2`
B. 1
C. `1/2`
D. `-1/2`

Answer

To determine the value of \( k \) for which the points \( (k, 2k) \), \( (2k, 3k) \), and \( (3, 1) \) are collinear, we can use the concept of the area of a triangle formed by these points. If the area is zero, then the points are collinear.

### Step-by-Step Solution:

1. **Set up the area formula**:
The area \( A \) of a triangle formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points, we have:
- \( (x_1, y_1) = (k, 2k) \)
- \( (x_2, y_2) = (2k, 3k) \)
- \( (x_3, y_3) = (3, 1) \)

2. **Substitute the points into the area formula**:
\[ A = \frac{1}{2} \left| k(3k - 1) + 2k(1 - 2k) + 3(2k - 3k) \right| \]

3. **Simplify the expression**:
Expanding the terms inside the absolute value:
- First term: \( k(3k - 1) = 3k^2 - k \)
- Second term: \( 2k(1 - 2k) = 2k - 4k^2 \)
- Third term: \( 3(2k - 3k) = 3(-k) = -3k \)

Combining these:
\[ A = \frac{1}{2} \left| 3k^2 - k + 2k - 4k^2 - 3k \right| \]
\[ = \frac{1}{2} \left| -k^2 - 2k \right| \]

4. **Set the area equal to zero**:
Since the points are collinear, the area must be zero:
\[ \frac{1}{2} \left| -k^2 - 2k \right| = 0 \]
This implies:
\[ -k^2 - 2k = 0 \]

5. **Factor the equation**:
Factoring out \( -k \):
\[ -k(k + 2) = 0 \]

6. **Find the values of \( k \)**:
Setting each factor to zero gives:
\[ k = 0 \quad \text{or} \quad k + 2 = 0 \Rightarrow k = -2 \]

7. **Conclusion**:
The possible values for \( k \) are \( 0 \) and \( -2 \). Since \( k = 0 \) does not provide a valid point in this context, we conclude:
\[ k = -2 \]

### Final Answer:
The value of \( k \) is \( -2 \).