The equation of a straight line on which the perpendicular from the origin of the length 2 units and this perpendicular makes an angle of `240^(@)` with the x-axis is (i) `x+sqrt(3)y+4=0` (ii) `x-sqrt(3)y+4=0` (iii) `x-sqrt(3)y-4=0` (iv) `x+sqrt(3)y-4=0`

Options

A. `x+sqrt(3)y+4=0`
B. `x-sqrt(3)y+4=0`
C. `x-sqrt(3)y-4=0`
D. `x+sqrt(3)y-4=0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We need to find the equation of a straight line where:
- A perpendicular from the origin has a length of 2 units.
- This perpendicular makes an angle of \(240^\circ\) with the x-axis.

### Step 2: Use the normal form of the line
The normal form of the equation of a line is given by:
\[ x \cos \theta + y \sin \theta = d \]
where:
- \(d\) is the length of the perpendicular from the origin (which is 2 in this case),
- \(\theta\) is the angle the perpendicular makes with the x-axis (which is \(240^\circ\)).

### Step 3: Substitute the values into the equation
Substituting \(d = 2\) and \(\theta = 240^\circ\):
\[ x \cos(240^\circ) + y \sin(240^\circ) = 2 \]

### Step 4: Calculate \(\cos(240^\circ)\) and \(\sin(240^\circ)\)
Using trigonometric values:
- \(\cos(240^\circ) = -\frac{1}{2}\)
- \(\sin(240^\circ) = -\frac{\sqrt{3}}{2}\)

### Step 5: Substitute the trigonometric values into the equation
Substituting these values into the equation gives:
\[ x \left(-\frac{1}{2}\right) + y \left(-\frac{\sqrt{3}}{2}\right) = 2 \]

### Step 6: Simplify the equation
Multiplying through by -2 to eliminate the fractions:
\[ x + \sqrt{3}y = -4 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x + \sqrt{3}y + 4 = 0 \]

### Step 8: Identify the correct option
Now we compare this with the given options:
- (i) \(x + \sqrt{3}y + 4 = 0\)
- (ii) \(x - \sqrt{3}y + 4 = 0\)
- (iii) \(x - \sqrt{3}y - 4 = 0\)
- (iv) \(x + \sqrt{3}y - 4 = 0\)

The correct option is (i) \(x + \sqrt{3}y + 4 = 0\).