The minimum value of p for which the lines `3x-4y=2, 3x-4y=12, 12x+5y=7` and `12x+5y=p` constitute the sides of a rhombus is

Options

A. 33
B. 19
C. `-19`
D. 9

Answer

To find the minimum value of \( p \) for which the lines \( 3x - 4y = 2 \), \( 3x - 4y = 12 \), \( 12x + 5y = 7 \), and \( 12x + 5y = p \) constitute the sides of a rhombus, we need to ensure that the distances between the pairs of parallel lines are equal.

### Step 1: Identify the lines and their parallel pairs
The given lines are:
1. \( L_1: 3x - 4y = 2 \)
2. \( L_2: 3x - 4y = 12 \)
3. \( L_3: 12x + 5y = 7 \)
4. \( L_4: 12x + 5y = p \)

The lines \( L_1 \) and \( L_2 \) are parallel because they have the same coefficients for \( x \) and \( y \). Similarly, \( L_3 \) and \( L_4 \) are parallel.

### Step 2: Calculate the distance between the parallel lines \( L_1 \) and \( L_2 \)
The formula for the distance \( d \) between two parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is given by:
\[ d = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} \]
For lines \( L_1 \) and \( L_2 \):
- \( a = 3 \), \( b = -4 \)
- \( c_1 = -2 \) (from \( 3x - 4y - 2 = 0 \))
- \( c_2 = -12 \) (from \( 3x - 4y - 12 = 0 \))

Calculating the distance:
\[ d_{12} = \frac{|(-12) - (-2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|-12 + 2|}{\sqrt{9 + 16}} = \frac{|-10|}{\sqrt{25}} = \frac{10}{5} = 2 \]

### Step 3: Calculate the distance between the parallel lines \( L_3 \) and \( L_4 \)
For lines \( L_3 \) and \( L_4 \):
- \( a = 12 \), \( b = 5 \)
- \( c_3 = -7 \) (from \( 12x + 5y - 7 = 0 \))
- \( c_4 = -p \) (from \( 12x + 5y - p = 0 \))

Calculating the distance:
\[ d_{34} = \frac{|(-p) - (-7)|}{\sqrt{12^2 + 5^2}} = \frac{|7 - p|}{\sqrt{144 + 25}} = \frac{|7 - p|}{\sqrt{169}} = \frac{|7 - p|}{13} \]

### Step 4: Set the distances equal for the rhombus condition
For the lines to form a rhombus, the distances must be equal:
\[ d_{12} = d_{34} \]
Substituting the distances:
\[ 2 = \frac{|7 - p|}{13} \]

### Step 5: Solve for \( p \)
Multiplying both sides by 13:
\[ 26 = |7 - p| \]

This absolute value equation gives us two cases:
1. \( 7 - p = 26 \)
2. \( 7 - p = -26 \)

**Case 1:**
\[ 7 - p = 26 \implies -p = 26 - 7 \implies -p = 19 \implies p = -19 \]

**Case 2:**
\[ 7 - p = -26 \implies -p = -26 - 7 \implies -p = -33 \implies p = 33 \]

### Step 6: Determine the minimum value of \( p \)
The two values we found for \( p \) are \( -19 \) and \( 33 \). The minimum value is:
\[ \text{Minimum value of } p = -19 \]

### Final Answer
The minimum value of \( p \) for which the lines constitute the sides of a rhombus is \( \boxed{-19} \).