Find the equation of straight lines through the point A(3,-2) and inclined at `60^(@)` to the line `sqrt(3)x+y=1`.

Answer

To find the equations of straight lines through the point A(3, -2) and inclined at \(60^\circ\) to the line \(\sqrt{3}x + y = 1\), we will follow these steps:

### Step 1: Find the slope of the given line
The equation of the line is given as:
\[ \sqrt{3}x + y = 1 \]
We can rewrite this in slope-intercept form \(y = mx + b\):
\[ y = -\sqrt{3}x + 1 \]
From this equation, we can see that the slope \(m_1\) of the given line is:
\[ m_1 = -\sqrt{3} \]

### Step 2: Use the angle of inclination to find the slopes of the required lines
Let \(m_2\) be the slope of the required lines. The angle between the two lines can be expressed using the formula:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Given that \(\theta = 60^\circ\), we know that:
\[ \tan 60^\circ = \sqrt{3} \]
Substituting \(m_1\) and \(\tan 60^\circ\) into the equation gives:
\[ \sqrt{3} = \frac{-\sqrt{3} - m_2}{1 - \sqrt{3} m_2} \]

### Step 3: Solve for \(m_2\)
Cross-multiplying gives:
\[ \sqrt{3}(1 - \sqrt{3} m_2) = -\sqrt{3} - m_2 \]
Expanding this, we get:
\[ \sqrt{3} - 3 m_2 = -\sqrt{3} - m_2 \]
Rearranging the equation:
\[ \sqrt{3} + \sqrt{3} = -m_2 + 3 m_2 \]
This simplifies to:
\[ 2\sqrt{3} = 2m_2 \implies m_2 = \sqrt{3} \]

Now, we also need to consider the negative case:
\[ \sqrt{3} = \frac{-\sqrt{3} - m_2}{1 - \sqrt{3} m_2} \]
Following similar steps will yield:
\[ \sqrt{3}(1 - \sqrt{3} m_2) = -\sqrt{3} - m_2 \]
This leads to:
\[ \sqrt{3} - 3 m_2 = -\sqrt{3} - m_2 \]
Rearranging gives:
\[ 2\sqrt{3} = -2m_2 \implies m_2 = -\sqrt{3} \]

Thus, the two possible slopes \(m_2\) are:
\[ m_2 = \sqrt{3} \quad \text{and} \quad m_2 = 0 \]

### Step 4: Find the equations of the lines
1. **For \(m_2 = \sqrt{3}\)**:
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(A(3, -2)\) and \(m_2 = \sqrt{3}\):
\[ y + 2 = \sqrt{3}(x - 3) \]
Rearranging gives:
\[ y + 2 = \sqrt{3}x - 3\sqrt{3} \implies \sqrt{3}x - y - 3\sqrt{3} + 2 = 0 \]

2. **For \(m_2 = 0\)**:
The line is horizontal:
\[ y + 2 = 0 \implies y = -2 \]

### Final Equations
The equations of the required lines are:
1. \(\sqrt{3}x - y - (3\sqrt{3} - 2) = 0\)
2. \(y + 2 = 0\)