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 follow these steps:
### Step 1: Write the equations in standard form
The equations of the lines are:
1. \( L_1: x + y - 1 = 0 \)
2. \( L_2: px + 2y + 1 = 0 \)
3. \( L_3: 4x + 2py + 7 = 0 \)
### Step 2: Set up the determinant for concurrency
For three lines to be concurrent, the determinant formed by their coefficients must equal zero. The determinant is given by:
\[
\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} = 0
\]
Where \( a_i, b_i, c_i \) are the coefficients of \( x, y, \) and the constant term respectively.
For our lines:
- From \( L_1 \): \( a_1 = 1, b_1 = 1, c_1 = -1 \)
- From \( L_2 \): \( a_2 = p, b_2 = 2, c_2 = 1 \)
- From \( L_3 \): \( a_3 = 4, b_3 = 2p, c_3 = 7 \)
### Step 3: Write the determinant
The determinant can be written as:
\[
\begin{vmatrix}
1 & 1 & -1 \\
p & 2 & 1 \\
4 & 2p & 7
\end{vmatrix}
= 0
\]
### Step 4: Calculate the determinant
Calculating the determinant:
\[
= 1 \begin{vmatrix}
2 & 1 \\
2p & 7
\end{vmatrix} - 1 \begin{vmatrix}
p & 1 \\
4 & 7
\end{vmatrix} - 1 \begin{vmatrix}
p & 2 \\
4 & 2p
\end{vmatrix}
\]
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:
\[
1(14 - 2p) - 1(7p - 4) - 1(2p^2 - 8) = 0
\]
### Step 5: Simplify the equation
This simplifies to:
\[
14 - 2p - 7p + 4 - 2p^2 + 8 = 0
\]
Combining like terms:
\[
-2p^2 - 9p + 26 = 0
\]
### Step 6: Rearranging the equation
Rearranging gives:
\[
2p^2 + 9p - 26 = 0
\]
### Step 7: Solve the quadratic equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 9, c = -26 \):
\[
p = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-26)}}{2 \cdot 2}
\]
Calculating the discriminant:
\[
= \sqrt{81 + 208} = \sqrt{289} = 17
\]
Thus,
\[
p = \frac{-9 \pm 17}{4}
\]
Calculating the two possible values:
1. \( p = \frac{8}{4} = 2 \)
2. \( p = \frac{-26}{4} = -\frac{13}{2} \)
### Step 8: Check for concurrency
We found two values of \( p \): \( 2 \) and \( -\frac{13}{2} \). However, we need to check if these values lead to concurrency.
For \( p = 2 \):
The line \( L_2 \) becomes \( 2x + 2y + 1 = 0 \), which has the same slope as \( L_1 \) (both have slope -1), thus they are parallel and cannot be concurrent.
For \( p = -\frac{13}{2} \):
This value does not lead to parallel lines, so it is valid.
### Conclusion
The only valid value of \( p \) for which the lines are concurrent is \( -\frac{13}{2} \).
Thus, the number of values of \( p \) for which the lines are concurrent is **1**.
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