If points (5,5), (10, k) and (-5, 1) are collinear, then the value of k is

Options

A. 8
B. 7
C. 9
D. 6

Answer

To determine the value of \( k \) such that the points \( (5, 5) \), \( (10, k) \), and \( (-5, 1) \) are collinear, we can use the concept of the area of a triangle formed by these three points. The area will be zero if the points are collinear.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We can use the determinant of a matrix formed by the coordinates of the points. The points can be represented as:
\[ (x_1, y_1) = (5, 5), \quad (x_2, y_2) = (10, k), \quad (x_3, y_3) = (-5, 1) \]
The determinant is given by:
\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]

2. **Substitute the Points into the Determinant**:
Substitute the coordinates into the determinant:
\[ \begin{vmatrix} 5 & 5 & 1 \\ 10 & k & 1 \\ -5 & 1 & 1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
Expanding the determinant, we have:
\[ 5 \begin{vmatrix} k & 1 \\ 1 & 1 \end{vmatrix} - 5 \begin{vmatrix} 10 & 1 \\ -5 & 1 \end{vmatrix} + 1 \begin{vmatrix} 10 & k \\ -5 & 1 \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
- First determinant:
\[ \begin{vmatrix} k & 1 \\ 1 & 1 \end{vmatrix} = k \cdot 1 - 1 \cdot 1 = k - 1 \]
- Second determinant:
\[ \begin{vmatrix} 10 & 1 \\ -5 & 1 \end{vmatrix} = 10 \cdot 1 - 1 \cdot (-5) = 10 + 5 = 15 \]
- Third determinant:
\[ \begin{vmatrix} 10 & k \\ -5 & 1 \end{vmatrix} = 10 \cdot 1 - k \cdot (-5) = 10 + 5k \]

4. **Combine the Results**:
Substitute back into the determinant equation:
\[ 5(k - 1) - 5(15) + (10 + 5k) = 0 \]
Simplifying this gives:
\[ 5k - 5 - 75 + 10 + 5k = 0 \]
Combine like terms:
\[ 10k - 70 = 0 \]

5. **Solve for \( k \)**:
Rearranging gives:
\[ 10k = 70 \implies k = 7 \]

### Final Answer:
The value of \( k \) is \( 7 \).