Sketch roughly the lines satisfying the given consition and write its equation angle of inclination = `150^(@)` and distance from the origin = 3 units .

Answer

To solve the problem of sketching the lines satisfying the given conditions (angle of inclination = 150° and distance from the origin = 3 units), we can follow these steps:

### Step 1: Understand the Given Information
- The angle of inclination (θ) is given as 150°.
- The distance from the origin is given as 3 units.

### Step 2: Find the Slope of the Line
The slope (m) of the line can be found using the tangent of the angle of inclination:
\[ m = \tan(θ) = \tan(150°) \]
Using the tangent function:
\[ \tan(150°) = \tan(180° - 30°) = -\tan(30°) = -\frac{1}{\sqrt{3}} \]
Thus, the slope \( m = -\frac{1}{\sqrt{3}} \).

### Step 3: Use the General Form of the Line Equation
The general equation of a line can be expressed as:
\[ y = mx + c \]
Substituting the slope we found:
\[ y = -\frac{1}{\sqrt{3}}x + c \]

### Step 4: Find the Distance from the Origin
The distance \( d \) from the origin to the line can be expressed as:
\[ d = \frac{|c|}{\sqrt{1 + m^2}} \]
We know that the distance \( d = 3 \):
\[ 3 = \frac{|c|}{\sqrt{1 + \left(-\frac{1}{\sqrt{3}}\right)^2}} \]
Calculating \( 1 + m^2 \):
\[ 1 + m^2 = 1 + \left(-\frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{4}{3} \]
Thus, \( \sqrt{1 + m^2} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \).

### Step 5: Substitute Back to Find \( c \)
Substituting back into the distance formula:
\[ 3 = \frac{|c|}{\frac{2}{\sqrt{3}}} \]
Cross-multiplying gives:
\[ 3 \cdot \frac{2}{\sqrt{3}} = |c| \]
Calculating:
\[ |c| = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]
Thus, \( c = \pm 2\sqrt{3} \).

### Step 6: Write the Equations of the Lines
Substituting \( c \) back into the line equation:
1. For \( c = 2\sqrt{3} \):
\[ y = -\frac{1}{\sqrt{3}}x + 2\sqrt{3} \]
Rearranging gives:
\[ x + \sqrt{3}y - 2\sqrt{3} = 0 \]

2. For \( c = -2\sqrt{3} \):
\[ y = -\frac{1}{\sqrt{3}}x - 2\sqrt{3} \]
Rearranging gives:
\[ x + \sqrt{3}y + 2\sqrt{3} = 0 \]

### Step 7: Sketch the Lines
To sketch the lines, we can find the intercepts:
- For \( x + \sqrt{3}y - 2\sqrt{3} = 0 \):
- Set \( y = 0 \) to find \( x \): \( x = 2\sqrt{3} \)
- Set \( x = 0 \) to find \( y \): \( y = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \)

- For \( x + \sqrt{3}y + 2\sqrt{3} = 0 \):
- Set \( y = 0 \) to find \( x \): \( x = -2\sqrt{3} \)
- Set \( x = 0 \) to find \( y \): \( y = -\frac{2\sqrt{3}}{\sqrt{3}} = -2 \)

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