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

Options

A. 3
B. 43892
C. 6
D. `(2sqrt(85))/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 point of the two adjacent sides
To find the intersection point of the lines \(4x + 5y = 0\) and \(7x + 2y = 0\), we can solve them simultaneously.

1. From \(4x + 5y = 0\), we can express \(y\) in terms of \(x\):
\[ y = -\frac{4}{5}x \]

2. Substitute \(y\) in the second equation \(7x + 2y = 0\):
\[ 7x + 2\left(-\frac{4}{5}x\right) = 0 \]
\[ 7x - \frac{8}{5}x = 0 \]
\[ \left(7 - \frac{8}{5}\right)x = 0 \]
\[ \frac{35}{5} - \frac{8}{5} = \frac{27}{5} \Rightarrow x = 0 \]

3. Substitute \(x = 0\) back into \(y = -\frac{4}{5}x\):
\[ y = 0 \]

Thus, the intersection point (origin) is \((0, 0)\).

### Step 2: Find the coordinates of the other vertices
Next, we need to find the coordinates of the vertices of the parallelogram. We will find the intersection of the diagonal \(11x + 7y = 9\) with the sides.

**Finding the intersection with \(4x + 5y = 0\)**:
1. Substitute \(y = -\frac{4}{5}x\) into \(11x + 7y = 9\):
\[ 11x + 7\left(-\frac{4}{5}x\right) = 9 \]
\[ 11x - \frac{28}{5}x = 9 \]
\[ \left(11 - \frac{28}{5}\right)x = 9 \]
\[ \frac{55}{5} - \frac{28}{5} = \frac{27}{5} \Rightarrow \frac{27}{5}x = 9 \Rightarrow x = \frac{9 \cdot 5}{27} = \frac{15}{27} = \frac{5}{9} \]

2. Substitute \(x = \frac{5}{9}\) back into \(y = -\frac{4}{5}x\):
\[ y = -\frac{4}{5}\left(\frac{5}{9}\right) = -\frac{4}{9} \]

So, one vertex is \(\left(\frac{5}{9}, -\frac{4}{9}\right)\).

**Finding the intersection with \(7x + 2y = 0\)**:
1. From \(7x + 2y = 0\), express \(y\) in terms of \(x\):
\[ y = -\frac{7}{2}x \]

2. Substitute into \(11x + 7y = 9\):
\[ 11x + 7\left(-\frac{7}{2}x\right) = 9 \]
\[ 11x - \frac{49}{2}x = 9 \]
\[ \left(11 - \frac{49}{2}\right)x = 9 \]
\[ \frac{22}{2} - \frac{49}{2} = -\frac{27}{2} \Rightarrow -\frac{27}{2}x = 9 \Rightarrow x = -\frac{9 \cdot 2}{27} = -\frac{6}{27} = -\frac{2}{9} \]

3. Substitute \(x = -\frac{2}{9}\) back into \(y = -\frac{7}{2}x\):
\[ y = -\frac{7}{2}\left(-\frac{2}{9}\right) = \frac{7}{9} \]

So, the second vertex is \(\left(-\frac{2}{9}, \frac{7}{9}\right)\).

### Step 3: Calculate the area of the parallelogram
The area of the parallelogram can be calculated using the formula for the area of a triangle formed by the vertices \((0, 0)\), \(\left(\frac{5}{9}, -\frac{4}{9}\right)\), and \(\left(-\frac{2}{9}, \frac{7}{9}\right)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0\left(-\frac{4}{9} - \frac{7}{9}\right) + \frac{5}{9}\left(\frac{7}{9} - 0\right) + \left(-\frac{2}{9}\right)(0 - (-\frac{4}{9})) \right| \]
\[ = \frac{1}{2} \left| 0 + \frac{5 \cdot 7}{81} + \frac{2 \cdot 4}{81} \right| \]
\[ = \frac{1}{2} \left| \frac{35 + 8}{81} \right| = \frac{1}{2} \left| \frac{43}{81} \right| = \frac{43}{162} \]

Since the area of the parallelogram is double that of the triangle:
\[ \text{Area of Parallelogram} = 2 \times \frac{43}{162} = \frac{43}{81} \]

### Final Answer
The area of the parallelogram is \(\frac{43}{81}\).