Find the angle between the following pairs of lines :
`(i) y=sqrt(3)x+1` and `y=(1)/(sqrt(3))x+2`
`(ii) y=x` and `y=1-x`
`(iii) 2x+3y=2` and `3x-2y=1`.

Answer

To find the angle between the given pairs of lines, we can use the formula:

\[ \tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} \]

where \( m_1 \) and \( m_2 \) are the slopes of the two lines.

### (i) Lines: \( y = \sqrt{3}x + 1 \) and \( y = \frac{1}{\sqrt{3}}x + 2 \)

1. **Identify the slopes**:
- For the first line, \( m_1 = \sqrt{3} \).
- For the second line, \( m_2 = \frac{1}{\sqrt{3}} \).

2. **Calculate \( \tan \theta \)**:
\[ \tan \theta = \frac{|\sqrt{3} - \frac{1}{\sqrt{3}}|}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} \]

3. **Simplify the numerator**:
- Convert \( \sqrt{3} \) to a common denominator:
\[ \sqrt{3} = \frac{3}{\sqrt{3}} \]
- Now, the numerator becomes:
\[ |\frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}}| = |\frac{3 - 1}{\sqrt{3}}| = \frac{2}{\sqrt{3}} \]

4. **Simplify the denominator**:
\[ 1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 + 1 = 2 \]

5. **Final calculation**:
\[ \tan \theta = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}} \]

6. **Find \( \theta \)**:
\[ \theta = 30^\circ \]

### (ii) Lines: \( y = x \) and \( y = 1 - x \)

1. **Identify the slopes**:
- For the first line, \( m_1 = 1 \).
- For the second line, \( m_2 = -1 \).

2. **Calculate \( \tan \theta \)**:
\[ \tan \theta = \frac{|1 - (-1)|}{1 + 1 \cdot (-1)} \]

3. **Simplify the numerator**:
\[ |1 + 1| = 2 \]

4. **Simplify the denominator**:
\[ 1 - 1 = 0 \]

5. **Final calculation**:
\[ \tan \theta = \frac{2}{0} \quad \text{(undefined)} \]

6. **Find \( \theta \)**:
\[ \theta = 90^\circ \]

### (iii) Lines: \( 2x + 3y = 2 \) and \( 3x - 2y = 1 \)

1. **Convert to slope-intercept form**:
- For \( 2x + 3y = 2 \):
\[ 3y = -2x + 2 \implies y = -\frac{2}{3}x + \frac{2}{3} \quad (m_1 = -\frac{2}{3}) \]
- For \( 3x - 2y = 1 \):
\[ -2y = -3x + 1 \implies y = \frac{3}{2}x - \frac{1}{2} \quad (m_2 = \frac{3}{2}) \]

2. **Calculate \( \tan \theta \)**:
\[ \tan \theta = \frac{|- \frac{2}{3} - \frac{3}{2}|}{1 + (-\frac{2}{3}) \cdot \frac{3}{2}} \]

3. **Simplify the numerator**:
- Find a common denominator:
\[ -\frac{2}{3} = -\frac{4}{6}, \quad \frac{3}{2} = \frac{9}{6} \]
- The numerator becomes:
\[ |-\frac{4}{6} - \frac{9}{6}| = |-\frac{13}{6}| = \frac{13}{6} \]

4. **Simplify the denominator**:
\[ 1 + (-1) = 0 \]

5. **Final calculation**:
\[ \tan \theta = \frac{\frac{13}{6}}{0} \quad \text{(undefined)} \]

6. **Find \( \theta \)**:
\[ \theta = 90^\circ \]

### Summary of Angles
- (i) \( 30^\circ \)
- (ii) \( 90^\circ \)
- (iii) \( 90^\circ \)