The value of k , if the point `A(2,3) B(4,k),C(6,-3)` are collinear is .

Options

A. 7
B. `-7`
C. 1
D. 0

Answer

To determine the value of \( k \) such that the points \( A(2,3) \), \( B(4,k) \), and \( C(6,-3) \) are collinear, we can use the condition for collinearity of three points. The points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) are collinear if the following determinant is equal to zero:

\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]

Substituting the coordinates of the points:

- \( A(2, 3) \) gives \( x_1 = 2 \), \( y_1 = 3 \)
- \( B(4, k) \) gives \( x_2 = 4 \), \( y_2 = k \)
- \( C(6, -3) \) gives \( x_3 = 6 \), \( y_3 = -3 \)

The determinant becomes:

\[ \begin{vmatrix} 2 & 3 & 1 \\ 4 & k & 1 \\ 6 & -3 & 1 \end{vmatrix} = 0 \]

Calculating the determinant:

\[ = 2 \begin{vmatrix} k & 1 \\ -3 & 1 \end{vmatrix} - 3 \begin{vmatrix} 4 & 1 \\ 6 & 1 \end{vmatrix} + 1 \begin{vmatrix} 4 & k \\ 6 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} k & 1 \\ -3 & 1 \end{vmatrix} = k \cdot 1 - 1 \cdot (-3) = k + 3 \)

2. \( \begin{vmatrix} 4 & 1 \\ 6 & 1 \end{vmatrix} = 4 \cdot 1 - 1 \cdot 6 = 4 - 6 = -2 \)

3. \( \begin{vmatrix} 4 & k \\ 6 & -3 \end{vmatrix} = 4 \cdot (-3) - k \cdot 6 = -12 - 6k \)

Substituting these back into the determinant:

\[ 2(k + 3) - 3(-2) + 1(-12 - 6k) = 0 \]

Expanding this:

\[ 2k + 6 + 6 - 12 - 6k = 0 \]

Combining like terms:

\[ -4k + 0 = 0 \]

This simplifies to:

\[ -4k = 0 \]

Thus, \( k = 0 \).

Therefore, the value of \( k \) is \( \boxed{0} \).