Reduce the following to the normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis
`(i) x-y=4` `(ii) y-2=0` `(iii) x-sqrt(3)y+8=0`

Answer

To solve the given equations and reduce them to normal form, we will follow a systematic approach for each equation. The normal form of a line is given by:

\[ x \cos \alpha + y \sin \alpha = p \]

where \( p \) is the perpendicular distance from the origin to the line, and \( \alpha \) is the angle between the perpendicular and the positive x-axis.

### (i) For the equation \( x - y = 4 \):

1. **Rearranging the equation**:
\[ x - y - 4 = 0 \]

2. **Identifying coefficients**:
Here, \( A = 1 \), \( B = -1 \), and \( C = -4 \).

3. **Finding \( p \)**:
The formula for the perpendicular distance \( p \) from the origin to the line \( Ax + By + C = 0 \) is:
\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|-4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

4. **Finding \( \alpha \)**:
\[ \tan \alpha = \frac{B}{A} = \frac{-1}{1} = -1 \implies \alpha = -\frac{\pi}{4} \text{ (or } 315^\circ\text{)} \]

5. **Normal form**:
The normal form is:
\[ x \cos\left(-\frac{\pi}{4}\right) + y \sin\left(-\frac{\pi}{4}\right) = 2\sqrt{2} \]

### (ii) For the equation \( y - 2 = 0 \):

1. **Rearranging the equation**:
\[ 0x + y - 2 = 0 \]

2. **Identifying coefficients**:
Here, \( A = 0 \), \( B = 1 \), and \( C = -2 \).

3. **Finding \( p \)**:
\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|-2|}{\sqrt{0^2 + 1^2}} = 2 \]

4. **Finding \( \alpha \)**:
Since \( A = 0 \) and \( B = 1 \):
\[ \alpha = \frac{\pi}{2} \text{ (or } 90^\circ\text{)} \]

5. **Normal form**:
The normal form is:
\[ 0 \cdot \cos\left(\frac{\pi}{2}\right) + y \sin\left(\frac{\pi}{2}\right) = 2 \]

### (iii) For the equation \( x - \sqrt{3}y + 8 = 0 \):

1. **Rearranging the equation**:
\[ x - \sqrt{3}y + 8 = 0 \]

2. **Identifying coefficients**:
Here, \( A = 1 \), \( B = -\sqrt{3} \), and \( C = 8 \).

3. **Finding \( p \)**:
\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|8|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{8}{\sqrt{4}} = 4 \]

4. **Finding \( \alpha \)**:
\[ \tan \alpha = \frac{B}{A} = \frac{-\sqrt{3}}{1} \implies \alpha = -\frac{\pi}{3} \text{ (or } 300^\circ\text{)} \]

5. **Normal form**:
The normal form is:
\[ x \cos\left(-\frac{\pi}{3}\right) + y \sin\left(-\frac{\pi}{3}\right) = 4 \]

### Summary of Results:
1. For \( x - y = 4 \):
- Normal form: \( x \cos\left(-\frac{\pi}{4}\right) + y \sin\left(-\frac{\pi}{4}\right) = 2\sqrt{2} \)
- \( p = 2\sqrt{2} \), \( \alpha = -\frac{\pi}{4} \)

2. For \( y - 2 = 0 \):
- Normal form: \( 0 \cdot \cos\left(\frac{\pi}{2}\right) + y \sin\left(\frac{\pi}{2}\right) = 2 \)
- \( p = 2 \), \( \alpha = \frac{\pi}{2} \)

3. For \( x - \sqrt{3}y + 8 = 0 \):
- Normal form: \( x \cos\left(-\frac{\pi}{3}\right) + y \sin\left(-\frac{\pi}{3}\right) = 4 \)
- \( p = 4 \), \( \alpha = -\frac{\pi}{3} \)