The angle between the lines x + y - 3 = 0 and x - y + 3 = 0 is `alpha` and the acute angle between the lines x - `sqrt3 y + 2 sqrt3 = 0` and `sqrt3 x - y + 1 = 0` is `beta` . Which one of the following is correct ?

Options

A. `alpha = beta`
B. `alpha gt beta`
C. `alpha lt beta`
D. `alpha = 2 beta`

Answer

To solve the problem, we need to find the angles between the given pairs of lines and then compare them.

### Step 1: Find the slopes of the first pair of lines
The first pair of lines is given by the equations:
1. \( x + y - 3 = 0 \)
2. \( x - y + 3 = 0 \)

We can rewrite these equations in slope-intercept form \( y = mx + b \).

For the first line:
\[ y = -x + 3 \quad \Rightarrow \quad \text{slope (m1)} = -1 \]

For the second line:
\[ y = x + 3 \quad \Rightarrow \quad \text{slope (m2)} = 1 \]

### Step 2: Calculate the angle \( \alpha \) between the first pair of lines
The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the slopes:
\[ \tan(\alpha) = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{0} \right| \]
Since the denominator is 0, this indicates that the lines are perpendicular, and thus:
\[ \alpha = 90^\circ \]

### Step 3: Find the slopes of the second pair of lines
The second pair of lines is given by:
1. \( x - \sqrt{3}y + 2\sqrt{3} = 0 \)
2. \( \sqrt{3}x - y + 1 = 0 \)

For the first line:
\[ \sqrt{3}y = x + 2\sqrt{3} \quad \Rightarrow \quad y = \frac{1}{\sqrt{3}}x + 2 \]
Thus, the slope (m3) is \( \frac{1}{\sqrt{3}} \).

For the second line:
\[ y = \sqrt{3}x + 1 \quad \Rightarrow \quad \text{slope (m4)} = \sqrt{3} \]

### Step 4: Calculate the angle \( \beta \) between the second pair of lines
Using the same formula:
\[ \tan(\beta) = \left| \frac{m_3 - m_4}{1 + m_3 m_4} \right| \]
Substituting the slopes:
\[ \tan(\beta) = \left| \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{1 + \left(\frac{1}{\sqrt{3}}\right)(\sqrt{3})} \right| = \left| \frac{\frac{1 - 3}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{-\frac{2}{\sqrt{3}}}{2} \right| = \frac{1}{\sqrt{3}} \]
Thus:
\[ \beta = 30^\circ \]

### Step 5: Compare \( \alpha \) and \( \beta \)
We have:
- \( \alpha = 90^\circ \)
- \( \beta = 30^\circ \)

Since \( \alpha > \beta \), the correct relationship is \( \alpha > \beta \).