The number of values of p for which the lines `x+ y - 1 = 0, px + 2y + 1 = 0` and `4x + 2py + 7 = 0` are concurrent is equal to

Options

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

Answer

To determine the number of values of \( p \) for which the lines \( x + y - 1 = 0 \), \( px + 2y + 1 = 0 \), and \( 4x + 2py + 7 = 0 \) are concurrent, we can use the condition for concurrency of three lines given by the determinant:

\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \]

Where the lines are represented in the form \( a_ix + b_iy + c_i = 0 \).

### Step 1: Identify coefficients
For the lines:
1. \( x + y - 1 = 0 \) can be written as \( 1x + 1y - 1 = 0 \) with \( a_1 = 1, b_1 = 1, c_1 = -1 \).
2. \( px + 2y + 1 = 0 \) gives \( a_2 = p, b_2 = 2, c_2 = 1 \).
3. \( 4x + 2py + 7 = 0 \) gives \( a_3 = 4, b_3 = 2p, c_3 = 7 \).

### Step 2: Set up the determinant
The determinant can be set up as follows:

\[ \begin{vmatrix} 1 & 1 & -1 \\ p & 2 & 1 \\ 4 & 2p & 7 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant using the formula for a 3x3 matrix:

\[ D = 1 \cdot \begin{vmatrix} 2 & 1 \\ 2p & 7 \end{vmatrix} - 1 \cdot \begin{vmatrix} p & 1 \\ 4 & 7 \end{vmatrix} - 1 \cdot \begin{vmatrix} p & 2 \\ 4 & 2p \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 1 \\ 2p & 7 \end{vmatrix} = (2)(7) - (1)(2p) = 14 - 2p \)
2. \( \begin{vmatrix} p & 1 \\ 4 & 7 \end{vmatrix} = (p)(7) - (1)(4) = 7p - 4 \)
3. \( \begin{vmatrix} p & 2 \\ 4 & 2p \end{vmatrix} = (p)(2p) - (2)(4) = 2p^2 - 8 \)

Putting it all together:

\[ D = 1(14 - 2p) - 1(7p - 4) - 1(2p^2 - 8) = 14 - 2p - 7p + 4 - 2p^2 + 8 \]

Combining like terms:

\[ D = -2p^2 - 9p + 26 \]

### Step 4: Set the determinant to zero
Setting the determinant equal to zero for concurrency:

\[ -2p^2 - 9p + 26 = 0 \]

### Step 5: Solve the quadratic equation
Multiplying through by -1 to simplify:

\[ 2p^2 + 9p - 26 = 0 \]

Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = 9, c = -26 \):

\[ p = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-26)}}{2 \cdot 2} \]
\[ = \frac{-9 \pm \sqrt{81 + 208}}{4} \]
\[ = \frac{-9 \pm \sqrt{289}}{4} \]
\[ = \frac{-9 \pm 17}{4} \]

Calculating the two possible values of \( p \):

1. \( p = \frac{8}{4} = 2 \)
2. \( p = \frac{-26}{4} = -\frac{13}{2} \)

### Step 6: Check for concurrency
Now we need to check if both values of \( p \) lead to concurrent lines.

For \( p = 2 \):
- The second line becomes \( 2x + 2y + 1 = 0 \), which is parallel to the first line \( x + y - 1 = 0 \) (since they have the same slope). Thus, they do not intersect.

For \( p = -\frac{13}{2} \):
- This value does not lead to parallel lines, and thus all three lines can be concurrent.

### Conclusion
Thus, the only value of \( p \) that allows the lines to be concurrent is \( -\frac{13}{2} \).

The number of values of \( p \) for which the lines are concurrent is:

\[ \boxed{1} \]