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

Options

A. `xsqrt2 + ysqrt3 =0`
B. `sqrt3 x - y = 4+3sqrt3`
C. `xsqrt3 +y = 3sqrt2 + 5`
D. none of (a) , (b) ,(c)

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 can be calculated using the formula:
\[ m = \tan(\theta) \]
Given that \( \theta = 60^\circ \):
\[ m = \tan(60^\circ) = \sqrt{3} \]

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

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

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

### Step 5: Write the equation in standard form
To express the equation in standard form, we can rearrange it:
\[ \sqrt{3}x - y - (3\sqrt{3} + 4) = 0 \]
This can be rewritten as:
\[ \sqrt{3}x - y = 3\sqrt{3} + 4 \]

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