For which value of k given below the point A (-1, 4), B (2, 5) and C (3, k) are collinear ?

Options

A. 16/3
B. 16
C. 5
D. -1

Answer

To determine the value of \( k \) for which the points \( A(-1, 4) \), \( B(2, 5) \), and \( C(3, k) \) are collinear, we can use the condition for collinearity of three points. The points are collinear if the following determinant is equal to zero:

\[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \]

### Step-by-Step Solution:

1. **Identify the Points:**
- Let \( A(-1, 4) \) be \( (x_1, y_1) \)
- Let \( B(2, 5) \) be \( (x_2, y_2) \)
- Let \( C(3, k) \) be \( (x_3, y_3) \)

Here, we have:
- \( x_1 = -1, y_1 = 4 \)
- \( x_2 = 2, y_2 = 5 \)
- \( x_3 = 3, y_3 = k \)

2. **Substitute the Values into the Collinearity Condition:**
\[ -1(5 - k) + 2(k - 4) + 3(4 - 5) = 0 \]

3. **Simplify the Expression:**
- First, calculate each term:
- \( -1(5 - k) = -5 + k \)
- \( 2(k - 4) = 2k - 8 \)
- \( 3(4 - 5) = 3(-1) = -3 \)

- Combine these:
\[ -5 + k + 2k - 8 - 3 = 0 \]

4. **Combine Like Terms:**
- Combine \( k \) terms and constant terms:
\[ (k + 2k) + (-5 - 8 - 3) = 0 \]
\[ 3k - 16 = 0 \]

5. **Solve for \( k \):**
- Add 16 to both sides:
\[ 3k = 16 \]
- Divide by 3:
\[ k = \frac{16}{3} \]

### Conclusion:
The value of \( k \) for which the points \( A(-1, 4) \), \( B(2, 5) \), and \( C(3, k) \) are collinear is:
\[ \boxed{\frac{16}{3}} \]