Convert the following equations into perpendicular form and find the length of perpendicular from origin and the angle between `x`-axis and the perpendicular from origin :
`(i) sqrt(3)x-y=8` `(ii) 2x+ysqrt(5)=6`

Answer

To solve the problem, we need to convert the given equations into their perpendicular form and then find the length of the perpendicular from the origin as well as the angle between the x-axis and the perpendicular from the origin.

### Step 1: Convert the first equation into perpendicular form

Given the equation:
\[ \sqrt{3}x - y = 8 \]

Rearranging it, we get:
\[ \sqrt{3}x - y - 8 = 0 \]

To convert this into the normal form \(x \cos \alpha + y \sin \alpha = p\), we need to express it in terms of \(x\) and \(y\) coefficients.

Divide the entire equation by 8:
\[ \frac{\sqrt{3}}{8}x - \frac{1}{8}y = 1 \]

Now, we can identify:
\[ \cos \alpha = \frac{\sqrt{3}}{8}, \quad \sin \alpha = -\frac{1}{8} \]

### Step 2: Find the angle \(\alpha\)

To find the angle \(\alpha\), we can use the tangent function:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{1}{8}}{\frac{\sqrt{3}}{8}} = -\frac{1}{\sqrt{3}} \]

This means:
\[ \alpha = 330^\circ \quad (\text{since } \tan \alpha < 0 \text{ in the fourth quadrant}) \]

### Step 3: Find the length of the perpendicular from the origin

The length of the perpendicular \(p\) from the origin to the line is given by:
\[ p = \frac{|d|}{\sqrt{a^2 + b^2}} \]
where \(d\) is the constant term (8 in our case), and \(a\) and \(b\) are the coefficients of \(x\) and \(y\) respectively.

Here, \(d = 8\), \(a = \sqrt{3}\), and \(b = -1\):
\[ p = \frac{|8|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = \frac{8}{\sqrt{3 + 1}} = \frac{8}{2} = 4 \]

### Step 4: Convert the second equation into perpendicular form

Given the second equation:
\[ 2x + y\sqrt{5} = 6 \]

Rearranging gives:
\[ 2x + y\sqrt{5} - 6 = 0 \]

Dividing the entire equation by 6:
\[ \frac{2}{6}x + \frac{\sqrt{5}}{6}y = 1 \]

Identifying the coefficients:
\[ \cos \alpha = \frac{1}{3}, \quad \sin \alpha = \frac{\sqrt{5}}{6} \]

### Step 5: Find the angle \(\alpha\) for the second equation

Using the tangent function:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{5}}{6}}{\frac{1}{3}} = \frac{\sqrt{5}}{2} \]

Thus:
\[ \alpha = \tan^{-1}\left(\frac{\sqrt{5}}{2}\right) \]

### Step 6: Find the length of the perpendicular from the origin for the second equation

Using the same formula for the length of the perpendicular:
\[ p = \frac{|d|}{\sqrt{a^2 + b^2}} \]
where \(d = 6\), \(a = 2\), and \(b = \sqrt{5}\):
\[ p = \frac{6}{\sqrt{2^2 + (\sqrt{5})^2}} = \frac{6}{\sqrt{4 + 5}} = \frac{6}{3} = 2 \]

### Final Results

1. For the first equation \(\sqrt{3}x - y = 8\):
- Length of the perpendicular from the origin: \(4\)
- Angle with the x-axis: \(330^\circ\)

2. For the second equation \(2x + y\sqrt{5} = 6\):
- Length of the perpendicular from the origin: \(2\)
- Angle with the x-axis: \(\tan^{-1}\left(\frac{\sqrt{5}}{2}\right)\)