The number of real values of k for which the lines `x-2y+3=0,kx+3y+1=0` and `4x-ky+2=0` are concurrent is:

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To find the number of real values of \( k \) for which the lines \( x - 2y + 3 = 0 \), \( kx + 3y + 1 = 0 \), and \( 4x - ky + 2 = 0 \) are concurrent, we will follow these steps:

### Step 1: Write the equations in standard form
The equations of the lines are:
1. \( x - 2y + 3 = 0 \)
2. \( kx + 3y + 1 = 0 \)
3. \( 4x - ky + 2 = 0 \)

### Step 2: Formulate the determinant
For the lines to be concurrent, the determinant of the coefficients of \( x \), \( y \), and the constant terms must be zero. The determinant can be expressed as:

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

### Step 3: Calculate the determinant
Calculating the determinant:

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

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 3 & -1 \\ -k & 2 \end{vmatrix} = 3(2) - (-1)(-k) = 6 - k \)
2. \( \begin{vmatrix} k & -1 \\ 4 & 2 \end{vmatrix} = k(2) - (-1)(4) = 2k + 4 \)
3. \( \begin{vmatrix} k & 3 \\ 4 & -k \end{vmatrix} = k(-k) - 3(4) = -k^2 - 12 \)

Substituting these back into the determinant:

\[ D = 1(6 - k) + 2(2k + 4) + 3(-k^2 - 12) \]

### Step 4: Simplify the determinant
Expanding the determinant:

\[ D = 6 - k + 4k + 8 - 3k^2 - 36 \]
\[ D = -3k^2 + 3k - 22 \]

### Step 5: Set the determinant to zero
Setting the determinant to zero for concurrency:

\[ -3k^2 + 3k - 22 = 0 \]

### Step 6: Solve the quadratic equation
We can rearrange this to:

\[ 3k^2 - 3k + 22 = 0 \]

### Step 7: Calculate the discriminant
The discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ D = b^2 - 4ac \]

Here, \( a = 3 \), \( b = -3 \), and \( c = 22 \):

\[ D = (-3)^2 - 4(3)(22) = 9 - 264 = -255 \]

### Step 8: Analyze the discriminant
Since the discriminant is negative (\( D < 0 \)), this means that the quadratic equation has no real solutions.

### Conclusion
Thus, the number of real values of \( k \) for which the lines are concurrent is **0**.

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