To solve the problem step by step, we will follow the given instructions and derive the necessary equations and area calculations.
### Step 1: Identify the pair of lines
The given equation is:
\[ 2x^2 - 5xy + 2y^2 = 0 \]
This represents a pair of straight lines. We need to factor this equation to find the equations of the lines.
### Step 2: Factor the quadratic equation
We can factor the quadratic using the method of splitting the middle term. The equation can be rewritten as:
\[ 2x^2 - 4xy - xy + 2y^2 = 0 \]
Grouping the terms, we get:
\[ (2x^2 - 4xy) + (-xy + 2y^2) = 0 \]
Factoring out common terms:
\[ 2x(x - 2y) - y(x - 2y) = 0 \]
This gives us:
\[ (2x - y)(x - 2y) = 0 \]
### Step 3: Write the equations of the lines
From the factorization, we have two lines:
1. \( 2x - y = 0 \) or \( y = 2x \) (Line 1)
2. \( x - 2y = 0 \) or \( y = \frac{1}{2}x \) (Line 2)
### Step 4: Identify the given diagonal
The diagonal of the parallelogram is given by the equation:
\[ 5x + 2y = 1 \]
### Step 5: Find the intersection points of the lines and the diagonal
To find the points of intersection, we will solve the equations of the lines with the diagonal.
#### Intersection with Line 1:
1. From \( y = 2x \), substitute into \( 5x + 2y = 1 \):
\[ 5x + 2(2x) = 1 \]
\[ 5x + 4x = 1 \]
\[ 9x = 1 \]
\[ x = \frac{1}{9} \]
\[ y = 2(\frac{1}{9}) = \frac{2}{9} \]
So, point A is \( \left(\frac{1}{9}, \frac{2}{9}\right) \).
#### Intersection with Line 2:
2. From \( y = \frac{1}{2}x \), substitute into \( 5x + 2y = 1 \):
\[ 5x + 2\left(\frac{1}{2}x\right) = 1 \]
\[ 5x + x = 1 \]
\[ 6x = 1 \]
\[ x = \frac{1}{6} \]
\[ y = \frac{1}{2}\left(\frac{1}{6}\right) = \frac{1}{12} \]
So, point C is \( \left(\frac{1}{6}, \frac{1}{12}\right) \).
### Step 6: Find the midpoint of diagonal AC
The midpoint H of diagonal AC can be calculated as:
\[ H = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) \]
\[ H = \left( \frac{\frac{1}{9} + \frac{1}{6}}{2}, \frac{\frac{2}{9} + \frac{1}{12}}{2} \right) \]
Calculating the x-coordinate:
\[ \frac{1}{9} + \frac{1}{6} = \frac{2}{18} + \frac{3}{18} = \frac{5}{18} \]
So,
\[ x_H = \frac{5/18}{2} = \frac{5}{36} \]
Calculating the y-coordinate:
\[ \frac{2}{9} + \frac{1}{12} = \frac{8}{36} + \frac{3}{36} = \frac{11}{36} \]
So,
\[ y_H = \frac{11/36}{2} = \frac{11}{72} \]
Thus, \( H = \left(\frac{5}{36}, \frac{11}{72}\right) \).
### Step 7: Find the equation of the other diagonal OB
The diagonal OB passes through the origin (0, 0) and point H. The slope \( m \) of line OB is given by:
\[ m = \frac{y_H - 0}{x_H - 0} = \frac{\frac{11}{72}}{\frac{5}{36}} = \frac{11}{72} \cdot \frac{36}{5} = \frac{11 \cdot 36}{72 \cdot 5} = \frac{11}{10} \]
Using the point-slope form of the line:
\[ y - 0 = \frac{11}{10}(x - 0) \]
Thus, the equation of the line is:
\[ y = \frac{11}{10}x \]
Rearranging gives:
\[ 11x - 10y = 0 \]
### Step 8: Calculate the area of the parallelogram
The area \( A \) of the parallelogram can be calculated using the formula:
\[ A = 2 \times \text{Area of triangle OAC} \]
Using the vertices O(0, 0), A\(\left(\frac{1}{9}, \frac{2}{9}\right)\), and C\(\left(\frac{1}{6}, \frac{1}{12}\right)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ = \frac{1}{2} \left| 0\left(\frac{2}{9} - \frac{1}{12}\right) + \frac{1}{9}\left(\frac{1}{12} - 0\right) + \frac{1}{6}\left(0 - \frac{2}{9}\right) \right| \]
Calculating:
\[ = \frac{1}{2} \left| 0 + \frac{1}{9} \cdot \frac{1}{12} - \frac{1}{6} \cdot \frac{2}{9} \right| \]
\[ = \frac{1}{2} \left| \frac{1}{108} - \frac{1}{27} \right| \]
Finding a common denominator (108):
\[ = \frac{1}{2} \left| \frac{1 - 4}{108} \right| = \frac{1}{2} \cdot \frac{3}{108} = \frac{1}{72} \]
Thus, the area of the parallelogram is:
\[ A = 2 \times \frac{1}{72} = \frac{1}{36} \]
### Final Answers
- The equation of the other diagonal is:
\[ 11x - 10y = 0 \]
- The area of the parallelogram is:
\[ \frac{1}{36} \text{ square units} \]
