Two adjacent sides of a parallelogram are `4x+5y=0`,and `7x+2y=0`. If the equation of one diagonal is `11x+7y=9`
Area of the parallelogram is

Options

A. A) 3
B. B) `3/2`
C. C) 6
D. D) `(2sqrt85)/(3)`

Answer

To find the area of the parallelogram formed by the lines \(4x + 5y = 0\) and \(7x + 2y = 0\) with one diagonal given by \(11x + 7y = 9\), we can follow these steps:

### Step 1: Find the intersection points of the lines

1. **Identify the equations of the sides and diagonal:**
- Side 1: \(4x + 5y = 0\) (Equation 1)
- Side 2: \(7x + 2y = 0\) (Equation 2)
- Diagonal: \(11x + 7y = 9\) (Equation 3)

2. **Find the intersection point of the two sides (A):**
- Solve the equations \(4x + 5y = 0\) and \(7x + 2y = 0\) simultaneously.
- From Equation 1, express \(y\) in terms of \(x\):
\[ y = -\frac{4}{5}x \]
- Substitute \(y\) into Equation 2:
\[ 7x + 2\left(-\frac{4}{5}x\right) = 0 \]
\[ 7x - \frac{8}{5}x = 0 \]
\[ \left(7 - \frac{8}{5}\right)x = 0 \]
\[ \left(\frac{35}{5} - \frac{8}{5}\right)x = 0 \]
\[ \frac{27}{5}x = 0 \implies x = 0 \]
- Substitute \(x = 0\) back into Equation 1 to find \(y\):
\[ 4(0) + 5y = 0 \implies y = 0 \]
- Thus, point A is \((0, 0)\).

### Step 2: Find the coordinates of points B and D

3. **Find point B (intersection of diagonal and side 1):**
- Substitute \(y = -\frac{4}{5}x\) into Equation 3:
\[ 11x + 7\left(-\frac{4}{5}x\right) = 9 \]
\[ 11x - \frac{28}{5}x = 9 \]
\[ \left(11 - \frac{28}{5}\right)x = 9 \]
\[ \left(\frac{55}{5} - \frac{28}{5}\right)x = 9 \]
\[ \frac{27}{5}x = 9 \implies x = \frac{9 \cdot 5}{27} = \frac{15}{27} = \frac{5}{9} \]
- Substitute \(x = \frac{5}{9}\) into \(y = -\frac{4}{5}x\):
\[ y = -\frac{4}{5}\left(\frac{5}{9}\right) = -\frac{4}{9} \]
- Thus, point B is \(\left(\frac{5}{9}, -\frac{4}{9}\right)\).

4. **Find point D (intersection of diagonal and side 2):**
- Substitute \(y = -\frac{7}{2}x\) into Equation 3:
\[ 11x + 7\left(-\frac{7}{2}x\right) = 9 \]
\[ 11x - \frac{49}{2}x = 9 \]
\[ \left(11 - \frac{49}{2}\right)x = 9 \]
\[ \left(\frac{22}{2} - \frac{49}{2}\right)x = 9 \]
\[ -\frac{27}{2}x = 9 \implies x = -\frac{9 \cdot 2}{27} = -\frac{18}{27} = -\frac{2}{3} \]
- Substitute \(x = -\frac{2}{3}\) into \(y = -\frac{7}{2}x\):
\[ y = -\frac{7}{2}\left(-\frac{2}{3}\right) = \frac{7}{3} \]
- Thus, point D is \(\left(-\frac{2}{3}, \frac{7}{3}\right)\).

### Step 3: Calculate the area of the parallelogram

5. **Use the area formula for a parallelogram given by vertices:**
- The area \(A\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Using points A \((0, 0)\), B \(\left(\frac{5}{9}, -\frac{4}{9}\right)\), and D \(\left(-\frac{2}{3}, \frac{7}{3}\right)\):
\[ A = \frac{1}{2} \left| 0\left(-\frac{4}{9} - \frac{7}{3}\right) + \frac{5}{9}\left(\frac{7}{3} - 0\right) + \left(-\frac{2}{3}\right)(0 - (-\frac{4}{9})) \right| \]
\[ = \frac{1}{2} \left| 0 + \frac{5}{9} \cdot \frac{7}{3} + \left(-\frac{2}{3}\right) \cdot \frac{4}{9} \right| \]
\[ = \frac{1}{2} \left| \frac{35}{27} - \frac{8}{27} \right| = \frac{1}{2} \left| \frac{27}{27} \right| = \frac{1}{2} \]

6. **Total area of the parallelogram:**
- Since the area of the triangle calculated is \(\frac{1}{2}\), the area of the parallelogram is:
\[ \text{Area} = 2 \times \frac{1}{2} = 1 \]

### Final Answer
The area of the parallelogram is \(1\) square unit.