Find the equation to the straight line passing through
the intersection of the lines
`x- 2y -a= 0 and x + 3y-2a = 0` and parallel to the straight line
`3x+ 4y= 0`

Answer

To find the equation of the straight line passing through the intersection of the lines \( x - 2y - a = 0 \) and \( x + 3y - 2a = 0 \), and parallel to the line \( 3x + 4y = 0 \), we can follow these steps:

### Step 1: Find the intersection of the two lines
To find the intersection point of the lines \( x - 2y - a = 0 \) and \( x + 3y - 2a = 0 \), we can solve these equations simultaneously.

1. From the first equation, express \( x \) in terms of \( y \):
\[ x = 2y + a \]

2. Substitute this expression for \( x \) into the second equation:
\[ (2y + a) + 3y - 2a = 0 \]
Simplifying this gives:
\[ 5y - a = 0 \implies y = \frac{a}{5} \]

3. Now substitute \( y = \frac{a}{5} \) back into the expression for \( x \):
\[ x = 2\left(\frac{a}{5}\right) + a = \frac{2a}{5} + \frac{5a}{5} = \frac{7a}{5} \]

Thus, the intersection point is \( \left(\frac{7a}{5}, \frac{a}{5}\right) \).

### Step 2: Write the general equation of the line through the intersection
The equation of a line passing through the intersection of the two given lines can be expressed as:
\[ x - 2y - a + \lambda (x + 3y - 2a) = 0 \]
where \( \lambda \) is a parameter.

### Step 3: Simplify the equation
Expanding the equation:
\[ x - 2y - a + \lambda x + 3\lambda y - 2a\lambda = 0 \]
Combining like terms:
\[ (1 + \lambda)x + (-2 + 3\lambda)y - (a + 2a\lambda) = 0 \]

### Step 4: Determine the slope of the line
The slope of the line in the form \( Ax + By + C = 0 \) is given by \( -\frac{A}{B} \). Here, \( A = 1 + \lambda \) and \( B = -2 + 3\lambda \). Thus, the slope is:
\[ \text{slope} = -\frac{1 + \lambda}{-2 + 3\lambda} = \frac{1 + \lambda}{2 - 3\lambda} \]

### Step 5: Set the slope equal to the slope of the line \( 3x + 4y = 0 \)
The slope of the line \( 3x + 4y = 0 \) is \( -\frac{3}{4} \). Setting the slopes equal gives:
\[ \frac{1 + \lambda}{2 - 3\lambda} = \frac{3}{4} \]

### Step 6: Cross-multiply and solve for \( \lambda \)
Cross-multiplying gives:
\[ 4(1 + \lambda) = 3(2 - 3\lambda) \]
Expanding both sides:
\[ 4 + 4\lambda = 6 - 9\lambda \]
Combining like terms:
\[ 4\lambda + 9\lambda = 6 - 4 \implies 13\lambda = 2 \implies \lambda = \frac{2}{13} \]

### Step 7: Substitute \( \lambda \) back into the equation
Substituting \( \lambda = \frac{2}{13} \) into the equation:
\[ (1 + \frac{2}{13})x + (-2 + 3 \cdot \frac{2}{13})y - \left(a + 2a \cdot \frac{2}{13}\right) = 0 \]
This simplifies to:
\[ \frac{15}{13}x + \left(-2 + \frac{6}{13}\right)y - \left(a + \frac{4a}{13}\right) = 0 \]
\[ \frac{15}{13}x + \left(-\frac{26}{13} + \frac{6}{13}\right)y - \frac{17a}{13} = 0 \]
\[ \frac{15}{13}x - \frac{20}{13}y - \frac{17a}{13} = 0 \]

### Step 8: Multiply through by 13 to eliminate the denominator
Multiplying through by 13 gives:
\[ 15x - 20y - 17a = 0 \]
Thus, the equation of the line is:
\[ 15x - 20y = 17a \]

### Final Answer
The equation of the straight line is:
\[ 15x - 20y = 17a \]