If the area of the triangle formed by a line with coordinates axes `54sqrt3`square units and the perpendicular drawn from the origin to the line makes an angle `60^(@)` with the x-axis, find the equation of the line.

Answer

To find the equation of the line that forms a triangle with the coordinate axes and has an area of \(54\sqrt{3}\) square units, we follow these steps:

### Step 1: Understand the Geometry
The line intersects the x-axis and y-axis at points A and B, forming a triangle OAB with the origin O. The area of triangle OAB is given as \(54\sqrt{3}\) square units.

### Step 2: Area of Triangle Formula
The area \(A\) of triangle OAB can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-intercept and the height is the y-intercept of the line.

### Step 3: Set Up the Line Equation
Assume the equation of the line is:
\[ y = mx + c \]
where \(m\) is the slope and \(c\) is the y-intercept.

### Step 4: Find the Intercepts
- The y-intercept (where the line intersects the y-axis) is \(c\).
- The x-intercept (where the line intersects the x-axis) can be found by setting \(y = 0\):
\[ 0 = mx + c \implies x = -\frac{c}{m} \]

### Step 5: Calculate the Area
Substituting the intercepts into the area formula:
\[ \text{Area} = \frac{1}{2} \times \left(-\frac{c}{m}\right) \times c = \frac{-c^2}{2m} \]
Setting this equal to the given area:
\[ \frac{-c^2}{2m} = 54\sqrt{3} \]
Thus,
\[ c^2 = -108m\sqrt{3} \]

### Step 6: Use the Angle Information
The perpendicular from the origin to the line makes an angle of \(60^\circ\) with the x-axis. The slope of this perpendicular line is:
\[ \tan(60^\circ) = \sqrt{3} \]
Let the slope of the line be \(m\). Since the two lines are perpendicular, we have:
\[ m \cdot (-\sqrt{3}) = -1 \implies m = \frac{1}{\sqrt{3}} \]

### Step 7: Substitute the Slope
Substituting \(m = \frac{1}{\sqrt{3}}\) into the equation for \(c^2\):
\[ c^2 = -108 \cdot \frac{1}{\sqrt{3}} \cdot \sqrt{3} = -108 \]
This indicates that \(c^2 = 108\), so:
\[ c = 6\sqrt{3} \quad \text{or} \quad c = -6\sqrt{3} \]

### Step 8: Write the Equations of the Line
Now substituting \(m\) and \(c\) back into the line equation:
1. For \(c = 6\sqrt{3}\):
\[ y = \frac{1}{\sqrt{3}}x + 6\sqrt{3} \]
2. For \(c = -6\sqrt{3}\):
\[ y = \frac{1}{\sqrt{3}}x - 6\sqrt{3} \]

### Step 9: Rearranging the Equation
We can rearrange the first equation:
\[ \sqrt{3}y = x + 18 \implies x - \sqrt{3}y + 18 = 0 \]
And for the second equation:
\[ \sqrt{3}y = x - 6\sqrt{3} \implies x - \sqrt{3}y - 6\sqrt{3} = 0 \]

### Final Result
The equations of the lines are:
1. \(x - \sqrt{3}y + 18 = 0\)
2. \(x - \sqrt{3}y - 6\sqrt{3} = 0\)