Statement-1: A circle can be inscribed in a quadrilateral whose sides are `3x -4y =0, 3x -4y= 5, 3x+ 4y= 0 and 3x+ 4y= 7` Statement-2: A circle can be inscribed in a parallelogram if and only if it is a rhombus

Options

A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
B. Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement-1.
C. Statement-1 is true, statement-2 is false.
D. Statement-1 is false, statement-2 is true.

Answer

To solve the given problem, we need to analyze both statements provided in the question.

### Step 1: Analyze Statement-1

The first statement claims that a circle can be inscribed in a quadrilateral whose sides are given by the equations:
1. \(3x - 4y = 0\)
2. \(3x - 4y = 5\)
3. \(3x + 4y = 0\)
4. \(3x + 4y = 7\)

To determine if a circle can be inscribed in this quadrilateral, we need to check if the sum of the lengths of the opposite sides is equal. A necessary and sufficient condition for a quadrilateral to have an inscribed circle is that the sum of the lengths of opposite sides must be equal.

### Step 2: Find the points of intersection

We will find the points of intersection of these lines to determine the vertices of the quadrilateral.

1. **Intersection of \(3x - 4y = 0\) and \(3x + 4y = 0\)**:
- Solve the equations:
\[ 3x - 4y = 0 \quad (1) \]
\[ 3x + 4y = 0 \quad (2) \]
- From (1), \(y = \frac{3}{4}x\).
- Substitute into (2):
\[ 3x + 4\left(\frac{3}{4}x\right) = 0 \Rightarrow 3x + 3x = 0 \Rightarrow 6x = 0 \Rightarrow x = 0 \Rightarrow y = 0 \]
- Point A: \((0, 0)\)

2. **Intersection of \(3x - 4y = 5\) and \(3x + 4y = 0\)**:
- Solve:
\[ 3x - 4y = 5 \quad (3) \]
\[ 3x + 4y = 0 \quad (2) \]
- From (2), \(y = -\frac{3}{4}x\).
- Substitute into (3):
\[ 3x - 4\left(-\frac{3}{4}x\right) = 5 \Rightarrow 3x + 3x = 5 \Rightarrow 6x = 5 \Rightarrow x = \frac{5}{6} \]
- Substitute \(x\) back to find \(y\):
\[ y = -\frac{3}{4}\left(\frac{5}{6}\right) = -\frac{15}{24} = -\frac{5}{8} \]
- Point B: \(\left(\frac{5}{6}, -\frac{5}{8}\right)\)

3. **Intersection of \(3x - 4y = 5\) and \(3x + 4y = 7\)**:
- Solve:
\[ 3x - 4y = 5 \quad (3) \]
\[ 3x + 4y = 7 \quad (4) \]
- Add (3) and (4):
\[ 6x = 12 \Rightarrow x = 2 \]
- Substitute \(x\) back to find \(y\):
\[ 3(2) - 4y = 5 \Rightarrow 6 - 4y = 5 \Rightarrow 4y = 1 \Rightarrow y = \frac{1}{4} \]
- Point C: \((2, \frac{1}{4})\)

4. **Intersection of \(3x - 4y = 0\) and \(3x + 4y = 7\)**:
- Solve:
\[ 3x - 4y = 0 \quad (1) \]
\[ 3x + 4y = 7 \quad (4) \]
- Add (1) and (4):
\[ 6x = 7 \Rightarrow x = \frac{7}{6} \]
- Substitute \(x\) back to find \(y\):
\[ 3\left(\frac{7}{6}\right) - 4y = 0 \Rightarrow \frac{7}{2} - 4y = 0 \Rightarrow 4y = \frac{7}{2} \Rightarrow y = \frac{7}{8} \]
- Point D: \(\left(\frac{7}{6}, \frac{7}{8}\right)\)

### Step 3: Calculate lengths of sides

Now, we will calculate the lengths of the sides formed by these points.

1. Length of side AB
2. Length of side BC
3. Length of side CD
4. Length of side DA

After calculating the lengths, we will check if the sum of opposite sides is equal.

### Step 4: Conclusion for Statement-1

If the sum of the lengths of opposite sides is equal, then a circle can be inscribed in the quadrilateral. If not, then it cannot.

### Step 5: Analyze Statement-2

The second statement claims that a circle can be inscribed in a parallelogram if and only if it is a rhombus. This is true because a rhombus has all sides equal, which is a necessary condition for a circle to be inscribed.

### Final Conclusion

- **Statement-1**: False (a circle cannot be inscribed in the given quadrilateral).
- **Statement-2**: True (only a rhombus can have a circle inscribed in it).