Find the equation of line through the intersection of lines `2x - y = 1 `and `3x + 2y = -9 ` and making an angle of `30^@` with positive direction of x-axis. 

Answer

To find the equation of the line that passes through the intersection of the lines \(2x - y = 1\) and \(3x + 2y = -9\) and makes an angle of \(30^\circ\) with the positive direction of the x-axis, we can follow these steps:

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

We have the equations:
1. \(2x - y = 1\) (Line 1)
2. \(3x + 2y = -9\) (Line 2)

To find the intersection, we can solve these equations simultaneously. We can express \(y\) from Line 1:

\[ y = 2x - 1 \]

Now, substitute \(y\) in Line 2:

\[ 3x + 2(2x - 1) = -9 \]

Expanding this gives:

\[ 3x + 4x - 2 = -9 \]

Combining like terms:

\[ 7x - 2 = -9 \]

Adding 2 to both sides:

\[ 7x = -7 \]

Dividing by 7:

\[ x = -1 \]

Now, substitute \(x = -1\) back into Line 1 to find \(y\):

\[ y = 2(-1) - 1 = -2 - 1 = -3 \]

Thus, the intersection point is \((-1, -3)\).

### Step 2: Determine the slope of the line

The angle given is \(30^\circ\). The slope \(m\) of a line making an angle \(\theta\) with the positive x-axis is given by:

\[ m = \tan(\theta) \]

For \(\theta = 30^\circ\):

\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 3: Use the point-slope form to find the equation of the line

We can use the point-slope form of the equation of a line, which is:

\[ y - y_1 = m(x - x_1) \]

Substituting \(m = \frac{1}{\sqrt{3}}\), \(x_1 = -1\), and \(y_1 = -3\):

\[ y - (-3) = \frac{1}{\sqrt{3}}(x - (-1)) \]

This simplifies to:

\[ y + 3 = \frac{1}{\sqrt{3}}(x + 1) \]

### Step 4: Rearranging the equation

Now, we can rearrange the equation:

\[ y + 3 = \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \]

Subtracting 3 from both sides:

\[ y = \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}} - 3 \]

### Final Equation

Thus, the equation of the line is:

\[ y = \frac{1}{\sqrt{3}}x + \left(\frac{1}{\sqrt{3}} - 3\right) \]