Find the area of the parallelogram formed by the lines
`2x^2+5xy+3y^2=0 and 2x^2+5xy+3y^2+3x+4y+1=0`

Answer

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

### Step 1: Identify the lines from the first equation
The first equation is:
\[ 2x^2 + 5xy + 3y^2 = 0 \]

We can factor this quadratic equation in terms of \(x\) and \(y\):
\[ 2x^2 + 5xy + 3y^2 = 0 \implies (2x + 3y)(x + y) = 0 \]
This gives us the two lines:
1. \( x + y = 0 \) (or \( y = -x \))
2. \( 2x + 3y = 0 \) (or \( y = -\frac{2}{3}x \))

### Step 2: Identify the lines from the second equation
The second equation is:
\[ 2x^2 + 5xy + 3y^2 + 3x + 4y + 1 = 0 \]

We can rearrange this as:
\[ 2x^2 + 5xy + 3y^2 = -3x - 4y - 1 \]
Since the left side is the same as the first equation, we can factor it similarly:
\[ (2x + 3y)(x + y) = -3x - 4y - 1 \]
We can assume the new lines are parallel to the previous lines:
1. \( x + y + 1 = 0 \) (or \( y = -x - 1 \))
2. \( 2x + 3y + 1 = 0 \) (or \( y = -\frac{2}{3}x - \frac{1}{3} \))

### Step 3: Identify coefficients for area calculation
Now we have four lines:
1. \( x + y = 0 \) (let's call this Line 1)
2. \( 2x + 3y = 0 \) (let's call this Line 2)
3. \( x + y + 1 = 0 \) (let's call this Line 3)
4. \( 2x + 3y + 1 = 0 \) (let's call this Line 4)

The coefficients we need for the area formula are:
- For Line 1: \( a_1 = 1, b_1 = 1, c_1 = 0 \)
- For Line 2: \( a_2 = 2, b_2 = 3, c_2 = 0 \)
- For Line 3: \( c_3 = 1 \)
- For Line 4: \( c_4 = 1 \)

### Step 4: Calculate the area of the parallelogram
The area \( A \) of the parallelogram formed by these lines can be calculated using the formula:
\[ A = \frac{|c_1 - c_2| \cdot |d_1 - d_2|}{|ad - bc|} \]
Where:
- \( c_1 = 1 \), \( c_2 = 0 \), \( d_1 = 0 \), \( d_2 = 1 \)
- \( a = 1 \), \( b = 1 \), \( c = 2 \), \( d = 3 \)

Substituting the values:
\[ A = \frac{|1 - 0| \cdot |0 - 1|}{|1 \cdot 3 - 1 \cdot 2|} = \frac{1 \cdot 1}{|3 - 2|} = \frac{1}{1} = 1 \]

### Final Answer
The area of the parallelogram is:
\[ \boxed{1} \]