For which value of `'p', y^(2)+xy+px^(2)-x-2y=0` represents a pair of straight lines

Options

A. `2`
B. `1/3`
C. `1/4`
D. `1/2`

Explanation

To determine the value of \( p \) for which the equation \( y^2 + xy + px^2 - x - 2y = 0 \) represents a pair of straight lines, we can follow these steps:

### Step 1: Rearrange the Equation
First, we rearrange the given equation into the standard form:
\[ px^2 + y^2 + xy - x - 2y = 0 \]

### Step 2: Identify Coefficients
Next, we identify the coefficients from the equation in the form \( ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \):
- \( a = p \)
- \( b = 1 \)
- \( 2h = 1 \) (thus \( h = \frac{1}{2} \))
- \( 2g = -1 \) (thus \( g = -\frac{1}{2} \))
- \( 2f = -2 \) (thus \( f = -1 \))
- \( c = 0 \)

### Step 3: Use the Condition for Pair of Straight Lines
For the equation to represent a pair of straight lines, the determinant \( \Delta \) must be equal to zero:
\[ \Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]

### Step 4: Substitute the Coefficients into the Determinant
Substituting the identified values into the determinant:
\[ \Delta = p \cdot 1 \cdot 0 + 2 \cdot (-1) \cdot \left(\frac{1}{2}\right) \cdot \left(-\frac{1}{2}\right) - p \cdot (-1)^2 - 1 \cdot \left(-\frac{1}{2}\right)^2 - 0 \cdot \left(\frac{1}{2}\right)^2 \]
This simplifies to:
\[ \Delta = 0 + 2 \cdot (-1) \cdot \frac{1}{4} - p - \frac{1}{4} - 0 \]
\[ \Delta = -\frac{1}{2} - p - \frac{1}{4} \]

### Step 5: Set the Determinant to Zero
Setting the determinant equal to zero gives:
\[ -\frac{1}{2} - p - \frac{1}{4} = 0 \]

### Step 6: Solve for \( p \)
Now, we solve for \( p \):
\[ -p = \frac{1}{2} + \frac{1}{4} \]
Converting \( \frac{1}{2} \) to quarters gives:
\[ -p = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \]
Thus,
\[ p = -\frac{3}{4} \]

### Conclusion
The value of \( p \) for which the equation represents a pair of straight lines is:
\[ \boxed{-\frac{3}{4}} \]