If the points A (1, -1) B ( 5, 2) and C (k , 5) are collinear, then `k=?`

Options

A. 2
B. 4
C. 6
D. 9

Answer

To find the value of \( k \) such that the points \( A(1, -1) \), \( B(5, 2) \), and \( C(k, 5) \) are collinear, we can use the formula for the area of a triangle formed by three points. If the area is zero, the points are collinear.

### Step-by-step Solution:

1. **Identify the coordinates of the points:**
- \( A(1, -1) \) corresponds to \( (x_1, y_1) \)
- \( B(5, 2) \) corresponds to \( (x_2, y_2) \)
- \( C(k, 5) \) corresponds to \( (x_3, y_3) \)

2. **Use the area formula for a triangle:**
The area \( A \) of a triangle formed by the points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Since the points are collinear, the area will be zero:
\[ \frac{1}{2} \left| 1(2 - 5) + 5(5 + 1) + k(-1 - 2) \right| = 0 \]

3. **Substitute the coordinates into the formula:**
\[ \frac{1}{2} \left| 1(-3) + 5(6) + k(-3) \right| = 0 \]
Simplifying this gives:
\[ \frac{1}{2} \left| -3 + 30 - 3k \right| = 0 \]

4. **Eliminate the fraction:**
Since the absolute value must equal zero, we can drop the \( \frac{1}{2} \):
\[ -3 + 30 - 3k = 0 \]

5. **Solve for \( k \):**
\[ 27 - 3k = 0 \]
Rearranging gives:
\[ 3k = 27 \]
Dividing both sides by 3:
\[ k = 9 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{9} \).