Find the slope and the inclination of the line AB if :
` A = (-1, 2sqrt(3))` and `B = (-2, sqrt(3))`

Answer

To find the slope and the inclination of the line AB given the points A = (-1, 2√3) and B = (-2, √3), we can follow these steps:

### Step 1: Identify the coordinates of points A and B
- Point A has coordinates \( A(-1, 2\sqrt{3}) \)
- Point B has coordinates \( B(-2, \sqrt{3}) \)

### 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 the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Here, we can assign:
- \( x_1 = -1 \), \( y_1 = 2\sqrt{3} \)
- \( x_2 = -2 \), \( y_2 = \sqrt{3} \)

### Step 3: Substitute the values into the slope formula
Substituting the values into the slope formula:
\[ m = \frac{\sqrt{3} - 2\sqrt{3}}{-2 - (-1)} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ \sqrt{3} - 2\sqrt{3} = -\sqrt{3} \]
Calculating the denominator:
\[ -2 + 1 = -1 \]
Now substituting back:
\[ m = \frac{-\sqrt{3}}{-1} = \sqrt{3} \]

### Step 5: Find the inclination angle
The inclination angle \( \theta \) can be found using the relationship between the slope and the tangent of the angle:
\[ m = \tan(\theta) \]
Since we found that \( m = \sqrt{3} \), we need to find \( \theta \) such that:
\[ \tan(\theta) = \sqrt{3} \]

### Step 6: Determine the angle
From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus, the inclination angle \( \theta \) is:
\[ \theta = 60^\circ \]

### Final Result
- The slope of line AB is \( \sqrt{3} \).
- The inclination of line AB is \( 60^\circ \).