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

Options

A. 0
B. `-9`
C. `-13`
D. 3

Answer

To determine the sum of all values of \( p \) for which the lines \( x + y - 1 = 0 \), \( px + 4y + 2 = 0 \), and \( 4x + py + 7 = 0 \) are concurrent, we will follow these steps:

### Step 1: Set up the equations
We have three lines represented by the equations:
1. \( x + y - 1 = 0 \) (Let this be Line 1)
2. \( px + 4y + 2 = 0 \) (Let this be Line 2)
3. \( 4x + py + 7 = 0 \) (Let this be Line 3)

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Line 1: \( a_1 = 1, b_1 = 1, c_1 = -1 \)
- For Line 2: \( a_2 = p, b_2 = 4, c_2 = 2 \)
- For Line 3: \( a_3 = 4, b_3 = p, c_3 = 7 \)

### Step 3: Set up the determinant for concurrency
The lines are concurrent if the determinant of the matrix formed by the coefficients is zero:
\[ \begin{vmatrix} 1 & 1 & -1 \\ p & 4 & 2 \\ 4 & p & 7 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ = 1 \cdot (4 \cdot 7 - 2 \cdot p) - 1 \cdot (p \cdot 7 - 2 \cdot 4) - 1 \cdot (p \cdot p - 4 \cdot 4) \]
\[ = 1 \cdot (28 - 2p) - 1 \cdot (7p - 8) - 1 \cdot (p^2 - 16) \]
\[ = 28 - 2p - 7p + 8 - p^2 + 16 \]
\[ = 52 - 9p - p^2 \]

### Step 5: Set the determinant to zero
Setting the determinant to zero gives:
\[ -p^2 - 9p + 52 = 0 \]
Rearranging gives:
\[ p^2 + 9p - 52 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 9, c = -52 \).
\[ p = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-52)}}{2 \cdot 1} \]
\[ = \frac{-9 \pm \sqrt{81 + 208}}{2} \]
\[ = \frac{-9 \pm \sqrt{289}}{2} \]
\[ = \frac{-9 \pm 17}{2} \]

Calculating the two possible values of \( p \):
1. \( p = \frac{8}{2} = 4 \)
2. \( p = \frac{-26}{2} = -13 \)

### Step 7: Check for concurrency
We need to check if \( p = 4 \) is valid. If \( p = 4 \):
- The equations \( 4x + 4y + 2 = 0 \) and \( 4x + 4y + 7 = 0 \) are inconsistent (same coefficients, different constants). Thus, they do not intersect.

So, \( p = 4 \) is not a valid solution.

### Step 8: Valid solution
The only valid solution is \( p = -13 \).

### Final Answer
The sum of all values of \( p \) for which the lines are concurrent is:
\[ \text{Sum} = -13 \]