The perimeter of a parallelogram whose sides are represented by the lines `x+2y+3=0`,
`3x+4y-5=0,2x+5=0 and 3x+4y-10=0` is equal to

Options

A. `(5)/(2)+5sqrt5` units
B. `5+4sqrt5` units
C. `5+(5)/(2)sqrt5` units
D. `(5+5sqrt5)/(2)` units

Answer

To find the perimeter of the parallelogram formed by the given lines, we will follow these steps:

### Step 1: Identify the equations of the lines
The given lines are:
1. \( x + 2y + 3 = 0 \)
2. \( 3x + 4y - 5 = 0 \)
3. \( 2x + 5 = 0 \)
4. \( 3x + 4y - 10 = 0 \)

### Step 2: Convert the equations to slope-intercept form
We will convert each line into the form \( y = mx + c \) to identify their slopes.

1. For \( x + 2y + 3 = 0 \):
\[ 2y = -x - 3 \implies y = -\frac{1}{2}x - \frac{3}{2} \]
(Slope \( m_1 = -\frac{1}{2} \))

2. For \( 3x + 4y - 5 = 0 \):
\[ 4y = -3x + 5 \implies y = -\frac{3}{4}x + \frac{5}{4} \]
(Slope \( m_2 = -\frac{3}{4} \))

3. For \( 2x + 5 = 0 \):
\[ 2x = -5 \implies x = -\frac{5}{2} \]
(This is a vertical line, slope is undefined)

4. For \( 3x + 4y - 10 = 0 \):
\[ 4y = -3x + 10 \implies y = -\frac{3}{4}x + \frac{10}{4} \]
(Slope \( m_3 = -\frac{3}{4} \))

### Step 3: Identify parallel lines
From the slopes, we can see:
- The lines \( 3x + 4y - 5 = 0 \) and \( 3x + 4y - 10 = 0 \) are parallel (same slope).
- The lines \( x + 2y + 3 = 0 \) and the vertical line \( 2x + 5 = 0 \) are not parallel to the above lines.

### Step 4: Calculate the distance between the parallel lines
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 the lines \( 3x + 4y - 5 = 0 \) and \( 3x + 4y - 10 = 0 \):
- Here, \( A = 3 \), \( B = 4 \), \( C_1 = -5 \), \( C_2 = -10 \).
\[ d_1 = \frac{|-10 + 5|}{\sqrt{3^2 + 4^2}} = \frac{5}{\sqrt{9 + 16}} = \frac{5}{5} = 1 \]

### Step 5: Calculate the distance between the other two lines
For the lines \( x + 2y + 3 = 0 \) and the vertical line \( 2x + 5 = 0 \):
- The distance from a point to a line can be used here. The point on the vertical line is \( (-\frac{5}{2}, y) \). We can find the distance from this point to the line \( x + 2y + 3 = 0 \).

Using the point-to-line distance formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Where \( A = 1, B = 2, C = 3 \), and \( (x_0, y_0) = (-\frac{5}{2}, y) \):
\[ d_2 = \frac{|1(-\frac{5}{2}) + 2y + 3|}{\sqrt{1^2 + 2^2}} = \frac{|-2.5 + 2y + 3|}{\sqrt{5}} = \frac{|2y + 0.5|}{\sqrt{5}} \]
To find \( y \), we can use the intersection of the two lines.

### Step 6: Find the perimeter
The perimeter \( P \) of a parallelogram is given by:
\[ P = 2(d_1 + d_2) \]
Substituting the values of \( d_1 \) and \( d_2 \) will give us the perimeter.

### Final Calculation
After calculating \( d_2 \) and substituting into the perimeter formula, we will arrive at the final answer.