If `(p+1) x+y=3 and 3y - (p-1)x=4` are perpendicular to each other, find the vlaue of p.

Answer

To find the value of \( p \) such that the lines represented by the equations \( (p+1)x + y = 3 \) and \( 3y - (p-1)x = 4 \) are perpendicular, we can follow these steps:

### Step 1: Rewrite the first equation in slope-intercept form
The first equation is:
\[ (p+1)x + y = 3 \]
Rearranging it to solve for \( y \):
\[ y = - (p + 1)x + 3 \]
From this, we can identify the slope \( m_1 \) of the first line:
\[ m_1 = - (p + 1) \]

### Step 2: Rewrite the second equation in slope-intercept form
The second equation is:
\[ 3y - (p - 1)x = 4 \]
Rearranging it to solve for \( y \):
\[ 3y = (p - 1)x + 4 \]
\[ y = \frac{(p - 1)}{3}x + \frac{4}{3} \]
From this, we can identify the slope \( m_2 \) of the second line:
\[ m_2 = \frac{(p - 1)}{3} \]

### Step 3: 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:
\[ - (p + 1) \cdot \frac{(p - 1)}{3} = -1 \]
Removing the negative signs:
\[ (p + 1) \cdot \frac{(p - 1)}{3} = 1 \]

### Step 4: Clear the fraction
Multiply both sides by 3 to eliminate the denominator:
\[ (p + 1)(p - 1) = 3 \]

### Step 5: Expand and rearrange the equation
Expanding the left-hand side:
\[ p^2 - 1 = 3 \]
Rearranging gives:
\[ p^2 - 1 - 3 = 0 \]
\[ p^2 - 4 = 0 \]

### Step 6: Factor the quadratic equation
Factoring gives:
\[ (p - 2)(p + 2) = 0 \]

### Step 7: Solve for \( p \)
Setting each factor to zero gives:
\[ p - 2 = 0 \quad \Rightarrow \quad p = 2 \]
\[ p + 2 = 0 \quad \Rightarrow \quad p = -2 \]

Thus, the values of \( p \) are:
\[ p = 2 \quad \text{or} \quad p = -2 \]