The side AB of an isosceles triangle is along the axis of x with vertices `A (–1, 0)` and `AB = AC`. The equation of the side BC when `angleA=120^(@)` and `BC=4sqrt(3)` is:

Options

A. `sqrt(3)x+y=3`
B. `x+y=sqrt(3)`
C. `x+sqrt(3)y=3`
D. None of these

Answer

To solve the problem, we need to find the equation of the line BC in the isosceles triangle ABC, where A is at (-1, 0), AB = AC, angle A = 120°, and BC = 4√3.

### Step-by-step Solution:

1. **Identify Points A and B**:
- Point A is given as A(-1, 0).
- Since AB is along the x-axis, we can denote point B as B(x_B, 0) where x_B > -1.

2. **Determine the Length of AB**:
- Let the length of AB be denoted as x. Therefore, the distance from A to B is:
\[ AB = |x_B - (-1)| = |x_B + 1| = x \]

3. **Determine the Length of AC**:
- Since AB = AC, we can denote the length of AC as x as well.

4. **Calculate the Angles**:
- Given that angle A = 120°, the angles at B and C will be equal since triangle ABC is isosceles. Therefore, the angles at B and C are:
\[ \text{Angle B} = \text{Angle C} = \frac{180° - 120°}{2} = 30° \]

5. **Find the Coordinates of Point C**:
- The angle at A is 120°, which means the angle at C is 30° with respect to the horizontal.
- The length of BC is given as 4√3.
- We can drop a perpendicular from point C to the x-axis, which we will denote as C'. The distance BC can be expressed in terms of the coordinates of C and C':
\[ BC = 4\sqrt{3} \]
- The coordinates of C can be expressed as C(x_C, y_C), where \( y_C = 4\sqrt{3} \sin(30°) = 4\sqrt{3} \cdot \frac{1}{2} = 2\sqrt{3} \).
- The horizontal distance (C'B) can be expressed as:
\[ C'B = 4\sqrt{3} \cos(30°) = 4\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 6 \]

6. **Calculate the x-coordinate of C**:
- Since B is at (x_B, 0) and the distance C'B is 6, we can express:
\[ C'B = |x_C - x_B| = 6 \]
- Thus, \( x_C = x_B + 6 \) or \( x_C = x_B - 6 \).

7. **Determine the Coordinates of Point B**:
- Since A is at (-1, 0) and the length AB is 4, we can find:
\[ x_B = -1 + 4 = 3 \]
- Therefore, B is at (3, 0).

8. **Find the Coordinates of Point C**:
- Using \( x_C = x_B + 6 \):
\[ x_C = 3 + 6 = 9 \]
- Thus, C is at (9, 2√3).

9. **Calculate the Slope of Line BC**:
- The slope of line BC can be calculated as:
\[ \text{slope} = \frac{y_C - y_B}{x_C - x_B} = \frac{2\sqrt{3} - 0}{9 - 3} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \]

10. **Write the Equation of Line BC**:
- Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (3, 0) \) and \( m = \frac{\sqrt{3}}{3} \):
\[ y - 0 = \frac{\sqrt{3}}{3}(x - 3) \]
Simplifying this gives:
\[ y = \frac{\sqrt{3}}{3}x - \sqrt{3} \]

### Final Answer:
The equation of line BC is:
\[ y = \frac{\sqrt{3}}{3}x - \sqrt{3} \]