Equation to the straight line cutting off an intercept 2 from negative direction of the axis of y and inclined at `30^@` to the positive direction of axis of x is :

Options

A. `y+x-sqrt(3)=0`
B. `y-x+2=0`
C. `y-xsqrt(3)-2=0`
D. `y sqrt(3)- x+2sqrt(3)=0`

Answer

To find the equation of the straight line that cuts off an intercept of 2 from the negative direction of the y-axis and is inclined at \(30^\circ\) to the positive direction of the x-axis, we can follow these steps:

### Step 1: Understand the given information
We know that:
- The y-intercept (c) is -2 (since it cuts off 2 from the negative direction of the y-axis).
- The angle of inclination (θ) is \(30^\circ\).

### Step 2: Calculate the slope (m) of the line
The slope \(m\) of a line can be calculated using the tangent of the angle of inclination:
\[ m = \tan(θ) = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 3: Write the equation of the line
The general equation of a line in slope-intercept form is given by:
\[ y = mx + c \]
Substituting the values of \(m\) and \(c\):
\[ y = \frac{1}{\sqrt{3}}x - 2 \]

### Step 4: Rearranging the equation
To express the equation in a standard form, we can multiply through by \(\sqrt{3}\) to eliminate the fraction:
\[ \sqrt{3}y = x - 2\sqrt{3} \]
Rearranging gives:
\[ \sqrt{3}y - x + 2\sqrt{3} = 0 \]

### Final Equation
Thus, the equation of the straight line is:
\[ \sqrt{3}y - x + 2\sqrt{3} = 0 \]

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