`3x^(2)+8xy-3y^(2)=0` represents a pair of lines AB and BC
`3x^(2)+8xy-3y^(2)+2x-4y-1=0` represents two lines CD and DA.

Options

A. The equation of CD is …………
B. The nature of the quadrilateral ABCD is ….
C. Area of quadrilaterla is…….
D. The coordinates of point D are……….

Answer

To solve the problem step by step, we will analyze the equations given and find the required lines and their intersections.

### Step 1: Factor the first equation
The first equation is given as:
\[ 3x^2 + 8xy - 3y^2 = 0 \]

We can factor this equation to find the lines it represents.

1. Rewrite the equation:
\[ 3x^2 + 8xy - 3y^2 = 0 \]

2. To factor, we can look for two numbers that multiply to \(3 \times -3 = -9\) and add to \(8\). These numbers are \(9\) and \(-1\).

3. Rewrite the middle term:
\[ 3x^2 + 9xy - xy - 3y^2 = 0 \]

4. Group the terms:
\[ (3x^2 + 9xy) + (-xy - 3y^2) = 0 \]

5. Factor by grouping:
\[ 3x(x + 3y) - y(x + 3y) = 0 \]

6. Factor out the common term:
\[ (3x - y)(x + 3y) = 0 \]

### Step 2: Identify the lines from the first equation
From the factorization, we get two lines:
1. \( 3x - y = 0 \) (Line AB)
2. \( x + 3y = 0 \) (Line BC)

### Step 3: Factor the second equation
The second equation is given as:
\[ 3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 \]

1. Combine like terms:
\[ 3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 \]

2. We will factor this equation similarly. We can rearrange it:
\[ 3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 \]

3. We can try to factor it directly or use the quadratic formula. Here, we will assume it can be factored as:
\[ (3x - y - 1)(x + 3y + 1) = 0 \]

### Step 4: Identify the lines from the second equation
From the factorization, we get two lines:
1. \( 3x - y - 1 = 0 \) (Line CD)
2. \( x + 3y + 1 = 0 \) (Line DA)

### Step 5: Find the intersection points
To find the intersection points of the lines, we can solve the equations pairwise.

#### Intersection of AB and BC
1. From \( 3x - y = 0 \) (AB), we have \( y = 3x \).
2. Substitute \( y \) in \( x + 3y = 0 \):
\[ x + 3(3x) = 0 \]
\[ x + 9x = 0 \]
\[ 10x = 0 \]
\[ x = 0 \]
Then, \( y = 3(0) = 0 \).
So, the intersection point A is \( (0, 0) \).

#### Intersection of CD and DA
1. From \( 3x - y - 1 = 0 \), we have \( y = 3x - 1 \).
2. Substitute \( y \) in \( x + 3y + 1 = 0 \):
\[ x + 3(3x - 1) + 1 = 0 \]
\[ x + 9x - 3 + 1 = 0 \]
\[ 10x - 2 = 0 \]
\[ 10x = 2 \]
\[ x = 0.2 \]
Then, \( y = 3(0.2) - 1 = 0.6 - 1 = -0.4 \).
So, the intersection point D is \( (0.2, -0.4) \).

### Summary of Results
- Lines AB and BC are represented by \( 3x - y = 0 \) and \( x + 3y = 0 \).
- Lines CD and DA are represented by \( 3x - y - 1 = 0 \) and \( x + 3y + 1 = 0 \).
- Intersection points are:
- A: \( (0, 0) \)
- D: \( (0.2, -0.4) \)