For what value of k, the points A (3, 4), B (5, k) and C (-3, 1) are collinear?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the value of \( k \) for which the points \( A(3, 4) \), \( B(5, k) \), and \( C(-3, 1) \) are collinear, we can use the concept that the area of the triangle formed by these three points must be zero.

### Step-by-Step Solution:

1. **Set Up the Area Formula:**
The area \( \Delta \) of the triangle formed by three points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points \( A(3, 4) \), \( B(5, k) \), and \( C(-3, 1) \), we can substitute:
- \( (x_1, y_1) = (3, 4) \)
- \( (x_2, y_2) = (5, k) \)
- \( (x_3, y_3) = (-3, 1) \)

2. **Substitute the Coordinates into the Formula:**
\[ \Delta = \frac{1}{2} \left| 3(k - 1) + 5(1 - 4) + (-3)(4 - k) \right| \]

3. **Simplify the Expression:**
\[ \Delta = \frac{1}{2} \left| 3k - 3 + 5(-3) - 3(4 - k) \right| \]
\[ = \frac{1}{2} \left| 3k - 3 - 15 - 12 + 3k \right| \]
\[ = \frac{1}{2} \left| 6k - 30 \right| \]

4. **Set the Area to Zero:**
For the points to be collinear, the area must be zero:
\[ \frac{1}{2} \left| 6k - 30 \right| = 0 \]
This implies:
\[ \left| 6k - 30 \right| = 0 \]

5. **Solve for \( k \):**
\[ 6k - 30 = 0 \]
\[ 6k = 30 \]
\[ k = 5 \]

### Final Answer:
The value of \( k \) for which the points \( A(3, 4) \), \( B(5, k) \), and \( C(-3, 1) \) are collinear is \( k = 5 \).