If `y+ (2p + 1) x + 3=0 and 8y- (2p-1)x=5` are mutually perpendicular, find the value of p.

Answer

To find the value of \( p \) such that the lines given by the equations \( y + (2p + 1)x + 3 = 0 \) and \( 8y - (2p - 1)x = 5 \) are mutually perpendicular, we can follow these steps:

### Step 1: Write the equations in slope-intercept form
The first equation is:
\[ y + (2p + 1)x + 3 = 0 \]
Rearranging gives:
\[ y = -(2p + 1)x - 3 \]
Thus, the slope \( m_1 \) of the first line is:
\[ m_1 = -(2p + 1) \]

The second equation is:
\[ 8y - (2p - 1)x = 5 \]
Rearranging gives:
\[ 8y = (2p - 1)x + 5 \quad \Rightarrow \quad y = \frac{(2p - 1)}{8}x + \frac{5}{8} \]
Thus, the slope \( m_2 \) of the second line is:
\[ m_2 = \frac{(2p - 1)}{8} \]

### Step 2: Use the condition for perpendicular lines
For two lines to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes we found:
\[ (-(2p + 1)) \cdot \left(\frac{(2p - 1)}{8}\right) = -1 \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{-(2p + 1)(2p - 1)}{8} = -1 \]
Multiplying both sides by \(-8\) gives:
\[ (2p + 1)(2p - 1) = 8 \]

### Step 4: Expand and rearrange
Expanding the left side:
\[ (2p)^2 - (1)^2 = 8 \quad \Rightarrow \quad 4p^2 - 1 = 8 \]
Adding \(1\) to both sides:
\[ 4p^2 = 9 \]

### Step 5: Solve for \( p \)
Dividing both sides by \(4\):
\[ p^2 = \frac{9}{4} \]
Taking the square root gives:
\[ p = \pm \frac{3}{2} \]

### Conclusion
Thus, the values of \( p \) that make the lines mutually perpendicular are:
\[ p = \frac{3}{2} \quad \text{or} \quad p = -\frac{3}{2} \]