Angle made by the line passsing through `(1,0) and (-2,sqrt3)` with x-axis is

Options

A. `120^(@)`
B. `150^(@)`
C. `60^(@)`
D. `135^(@)`

Answer

To find the angle made by the line passing through the points (1, 0) and (-2, √3) with the x-axis, we can follow these steps:

### Step 1: Identify the coordinates of the points
Let the points be:
- \( P_1(1, 0) \) and \( P_2(-2, \sqrt{3}) \)

### Step 2: Use the slope formula
The slope \( m \) of the line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates:
- \( x_1 = 1, y_1 = 0 \)
- \( x_2 = -2, y_2 = \sqrt{3} \)

We can calculate:
\[ m = \frac{\sqrt{3} - 0}{-2 - 1} = \frac{\sqrt{3}}{-3} = -\frac{\sqrt{3}}{3} \]

### Step 3: Relate the slope to the angle
The angle \( \theta \) that the line makes with the positive x-axis is given by:
\[ \tan(\theta) = m \]
Thus, we have:
\[ \tan(\theta) = -\frac{\sqrt{3}}{3} \]

### Step 4: Find the angle using the inverse tangent function
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) \]
We know that:
\[ \tan\left(30^\circ\right) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan\left(-30^\circ\right) = -\frac{1}{\sqrt{3}} \]
Thus:
\[ \theta = -30^\circ \]

### Step 5: Adjust the angle to find the positive equivalent
Since angles are often expressed in the range of \( 0^\circ \) to \( 360^\circ \), we can convert \( -30^\circ \) to a positive angle:
\[ \theta = 360^\circ - 30^\circ = 330^\circ \]
However, we also note that the angle can be expressed in the second quadrant:
\[ \theta = 180^\circ - 30^\circ = 150^\circ \]

### Final Answer
Thus, the angle made by the line with the positive x-axis is:
\[ \theta = 150^\circ \]

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