The equation `2x^(2)-3xy-py^(2)+x+qy-1=0` represent two mutually perpendicular lines if

Options

A. `p=3, q=2`
B. `p=2, q=3`
C. `p=-2, q=3`
D. `p=2, q=-9//2.`

Explanation

To determine the conditions under which the equation \(2x^2 - 3xy - py^2 + x + qy - 1 = 0\) represents two mutually perpendicular lines, we will follow these steps:

### Step 1: Identify coefficients
The general form of the equation for a pair of straight lines is given by:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify the coefficients:
- \(a = 2\)
- \(h = -\frac{3}{2}\) (since the coefficient of \(xy\) is \(-3\))
- \(b = -p\)
- \(g = \frac{1}{2}\) (since \(2g = 1\))
- \(f = \frac{q}{2}\) (since \(2f = q\))
- \(c = -1\)

### Step 2: Condition for perpendicular lines
For the lines to be mutually perpendicular, the following condition must hold:
\[ a + b = 0 \]
Substituting the values we have:
\[ 2 - p = 0 \implies p = 2 \]

### Step 3: Use the second condition
Next, we use the condition for the pair of straight lines:
\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]
Substituting the values:
- \(a = 2\)
- \(b = -p = -2\)
- \(c = -1\)
- \(g = \frac{1}{2}\)
- \(f = \frac{q}{2}\)
- \(h = -\frac{3}{2}\)

Now, substituting these into the condition:
\[ (2)(-2)(-1) + 2\left(\frac{q}{2}\right)\left(-\frac{3}{2}\right)\left(\frac{1}{2}\right) - (2)\left(\frac{q}{2}\right)^2 - (-2)\left(\frac{1}{2}\right)^2 - (-1)\left(-\frac{3}{2}\right)^2 = 0 \]

### Step 4: Simplifying the equation
Calculating each term:
1. \(abc = 2 \cdot (-2) \cdot (-1) = 4\)
2. \(2fgh = 2 \cdot \frac{q}{2} \cdot -\frac{3}{2} \cdot \frac{1}{2} = -\frac{3q}{2}\)
3. \(-af^2 = -2 \cdot \left(\frac{q}{2}\right)^2 = -\frac{q^2}{2}\)
4. \(-bg^2 = -(-2) \cdot \left(\frac{1}{2}\right)^2 = -\frac{1}{2}\)
5. \(-ch^2 = -(-1) \cdot \left(-\frac{3}{2}\right)^2 = -\frac{9}{4}\)

Combining these:
\[ 4 - \frac{3q}{2} - \frac{q^2}{2} - \frac{1}{2} - \frac{9}{4} = 0 \]

### Step 5: Combine and simplify
Combining the constants:
\[ 4 - \frac{1}{2} - \frac{9}{4} = \frac{16}{4} - \frac{1}{2} - \frac{9}{4} = \frac{16 - 2 - 9}{4} = \frac{5}{4} \]
Thus, we have:
\[ \frac{5}{4} - \frac{3q}{2} - \frac{q^2}{2} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 5 - 6q - 2q^2 = 0 \]
Rearranging gives:
\[ 2q^2 + 6q - 5 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:
\[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 + 40}}{4} = \frac{-6 \pm \sqrt{76}}{4} = \frac{-6 \pm 2\sqrt{19}}{4} = \frac{-3 \pm \sqrt{19}}{2} \]

### Final Result
Thus, the values of \(q\) that satisfy the condition for the lines to be mutually perpendicular are:
\[ q = \frac{-3 + \sqrt{19}}{2} \quad \text{or} \quad q = \frac{-3 - \sqrt{19}}{2} \]