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 (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

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 problem, we need to analyze the two statements given and determine their validity.

### Step 1: Analyze Statement-1
**Statement-1:** 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 quadrilateral can have an inscribed circle (is tangential) if and only if the sum of the lengths of opposite sides is equal.

### Step 2: Find the vertices of the quadrilateral
To find the vertices, we need to solve the equations pairwise:

1. **Intersection of \(3x - 4y = 0\) and \(3x + 4y = 0\):**
- From \(3x - 4y = 0\), we have \(y = \frac{3}{4}x\).
- Substitute into \(3x + 4y = 0\):
\[ 3x + 4\left(\frac{3}{4}x\right) = 0 \implies 3x + 3x = 0 \implies x = 0 \implies y = 0. \]
- Vertex A: \((0, 0)\).

2. **Intersection of \(3x - 4y = 0\) and \(3x + 4y = 7\):**
- From \(3x - 4y = 0\), \(y = \frac{3}{4}x\).
- Substitute into \(3x + 4y = 7\):
\[ 3x + 4\left(\frac{3}{4}x\right) = 7 \implies 3x + 3x = 7 \implies 6x = 7 \implies x = \frac{7}{6}, y = \frac{3}{8}. \]
- Vertex B: \(\left(\frac{7}{6}, \frac{3}{8}\right)\).

3. **Intersection of \(3x - 4y = 5\) and \(3x + 4y = 0\):**
- From \(3x + 4y = 0\), \(y = -\frac{3}{4}x\).
- Substitute into \(3x - 4y = 5\):
\[ 3x - 4\left(-\frac{3}{4}x\right) = 5 \implies 3x + 3x = 5 \implies 6x = 5 \implies x = \frac{5}{6}, y = -\frac{5}{8}. \]
- Vertex C: \(\left(\frac{5}{6}, -\frac{5}{8}\right)\).

4. **Intersection of \(3x - 4y = 5\) and \(3x + 4y = 7\):**
- From \(3x + 4y = 7\), \(y = \frac{7 - 3x}{4}\).
- Substitute into \(3x - 4y = 5\):
\[ 3x - 4\left(\frac{7 - 3x}{4}\right) = 5 \implies 3x - (7 - 3x) = 5 \implies 6x - 7 = 5 \implies 6x = 12 \implies x = 2, y = -\frac{1}{2}. \]
- Vertex D: \((2, -\frac{1}{2})\).

### Step 3: Calculate the lengths of the sides
Now we need to find the lengths of the sides using the distance formula between the vertices we found.

1. Length of side AB:
\[ AB = \sqrt{\left(\frac{7}{6} - 0\right)^2 + \left(\frac{3}{8} - 0\right)^2} \]

2. Length of side BC:
\[ BC = \sqrt{\left(\frac{5}{6} - \frac{7}{6}\right)^2 + \left(-\frac{5}{8} - \frac{3}{8}\right)^2} \]

3. Length of side CD:
\[ CD = \sqrt{\left(2 - \frac{5}{6}\right)^2 + \left(-\frac{1}{2} - (-\frac{5}{8})\right)^2} \]

4. Length of side DA:
\[ DA = \sqrt{\left(0 - 2\right)^2 + \left(0 - (-\frac{1}{2})\right)^2} \]

### Step 4: Check if the quadrilateral is tangential
After calculating the lengths, we check if \(AB + CD = BC + DA\). If they are equal, Statement-1 is true; otherwise, it is false.

### Step 5: Analyze Statement-2
**Statement-2:** A circle can be inscribed in a parallelogram if and only if it is a rhombus.

This statement is true because only in a rhombus are the opposite sides equal and the angles are equal, allowing for a circle to be inscribed.

### Conclusion
After analyzing both statements:
- Statement-1 is false (a circle cannot be inscribed in the given quadrilateral).
- Statement-2 is true (a circle can be inscribed in a rhombus).

Thus, the correct answer is **(d) Statement-1 is false, Statement-2 is true.**