Find the equation of the line which passes through the point `(3,-4)` and makes an angle of `60^(@)` with the positive direction of x - axis ?

Options

A. `xsqrt(2) + y sqrt(3) = 0`
B. `sqrt(3) x - y = 4+3 sqrt(3)`
C. `xsqrt(3)+y = 3 sqrt(2) +5`
D. None of these

Answer

To find the equation of the line that passes through the point (3, -4) and makes an angle of 60 degrees with the positive direction of the x-axis, we can follow these steps:

### Step 1: Determine the slope of the line
The slope \( m \) of a line that makes an angle \( \theta \) with the positive x-axis is given by:
\[ m = \tan(\theta) \]
For \( \theta = 60^\circ \):
\[ m = \tan(60^\circ) = \sqrt{3} \]

### Step 2: Use the point-slope form of the equation of a line
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Here, \( (x_1, y_1) = (3, -4) \) and \( m = \sqrt{3} \).

### Step 3: Substitute the values into the point-slope form
Substituting the values into the equation:
\[ y - (-4) = \sqrt{3}(x - 3) \]
This simplifies to:
\[ y + 4 = \sqrt{3}(x - 3) \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to the standard form:
\[ y + 4 = \sqrt{3}x - 3\sqrt{3} \]
Subtracting \( 4 \) from both sides gives:
\[ y = \sqrt{3}x - 3\sqrt{3} - 4 \]

### Step 5: Rearranging to standard form
To express it in the standard form \( Ax + By + C = 0 \):
\[ -\sqrt{3}x + y + (4 + 3\sqrt{3}) = 0 \]
Multiplying through by -1 to make the coefficient of \( x \) positive:
\[ \sqrt{3}x - y - (4 + 3\sqrt{3}) = 0 \]

### Final Equation
Thus, the equation of the line is:
\[ \sqrt{3}x - y = 4 + 3\sqrt{3} \]