Equation(s) of the straight line(s), inclined at `30^0` to the x-axis such that the length of its (each of their) line segment(s) between the coordinate axes is 10 units, is (are) `x+sqrt(3)y+5sqrt(3)=0` `x-sqrt(3)y+5sqrt(3)=0` `x+sqrt(3)y-5sqrt(3)=0` `x-sqrt(3)y-5sqrt(3)=0`

Options

A. `x+sqrt(3)y+5sqrt(3)=0`
B. `x-sqrt(3)y+5sqrt(3)=0`
C. `x+sqrt(3)y-5sqrt(3)=0`
D. `x-sqrt(3)y-5sqrt(3)=0`

Explanation

To find the equation of the straight lines inclined at \(30^\circ\) to the x-axis such that the length of their line segment between the coordinate axes is 10 units, we can follow these steps:

### Step 1: Understand the Geometry
The line makes an angle of \(30^\circ\) with the x-axis. The length of the line segment between the x-axis and y-axis is given as 10 units.

**Hint:** Recall that the angle with the x-axis will help us determine the slope of the line.

### Step 2: Determine the Slope
The slope \(m\) of a line inclined at an angle \(\theta\) with the x-axis is given by:
\[ m = \tan(\theta) \]
For \(\theta = 30^\circ\):
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

**Hint:** The slope is crucial for writing the equation of the line.

### Step 3: Use the Length of the Line Segment
Let the x-intercept be \(a\) and the y-intercept be \(b\). The length of the line segment between the axes can be expressed as:
\[ \text{Length} = \sqrt{a^2 + b^2} = 10 \]
Squaring both sides gives:
\[ a^2 + b^2 = 100 \]

**Hint:** This equation relates the x-intercept and y-intercept.

### Step 4: Express the Intercepts in Terms of the Slope
From the slope \(m = \frac{b}{a}\), we can express \(b\) in terms of \(a\):
\[ b = \frac{1}{\sqrt{3}} a \]
Substituting \(b\) into the length equation:
\[ a^2 + \left(\frac{1}{\sqrt{3}} a\right)^2 = 100 \]
This simplifies to:
\[ a^2 + \frac{1}{3} a^2 = 100 \]
\[ \frac{4}{3} a^2 = 100 \]
\[ a^2 = 75 \quad \Rightarrow \quad a = \pm 5\sqrt{3} \]

**Hint:** Solving for \(a\) gives us the x-intercepts.

### Step 5: Find Corresponding y-intercepts
Using \(a = 5\sqrt{3}\):
\[ b = \frac{1}{\sqrt{3}} (5\sqrt{3}) = 5 \]
Using \(a = -5\sqrt{3}\):
\[ b = \frac{1}{\sqrt{3}} (-5\sqrt{3}) = -5 \]

**Hint:** Each x-intercept corresponds to a y-intercept.

### Step 6: Write the Equations of the Lines
Now we have two pairs of intercepts:
1. \( (5\sqrt{3}, 5) \)
2. \( (5\sqrt{3}, -5) \)
3. \( (-5\sqrt{3}, 5) \)
4. \( (-5\sqrt{3}, -5) \)

Using the intercept form of the line equation:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
We can write the equations:
1. For \( (5\sqrt{3}, 5) \):
\[ \frac{x}{5\sqrt{3}} + \frac{y}{5} = 1 \quad \Rightarrow \quad x + \sqrt{3}y - 5\sqrt{3} = 0 \]
2. For \( (5\sqrt{3}, -5) \):
\[ \frac{x}{5\sqrt{3}} + \frac{y}{-5} = 1 \quad \Rightarrow \quad x - \sqrt{3}y - 5\sqrt{3} = 0 \]
3. For \( (-5\sqrt{3}, 5) \):
\[ \frac{x}{-5\sqrt{3}} + \frac{y}{5} = 1 \quad \Rightarrow \quad x + \sqrt{3}y + 5\sqrt{3} = 0 \]
4. For \( (-5\sqrt{3}, -5) \):
\[ \frac{x}{-5\sqrt{3}} + \frac{y}{-5} = 1 \quad \Rightarrow \quad x - \sqrt{3}y + 5\sqrt{3} = 0 \]

### Final Answer
The equations of the required lines are:
1. \( x + \sqrt{3}y - 5\sqrt{3} = 0 \)
2. \( x - \sqrt{3}y - 5\sqrt{3} = 0 \)
3. \( x + \sqrt{3}y + 5\sqrt{3} = 0 \)
4. \( x - \sqrt{3}y + 5\sqrt{3} = 0 \)