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 an isosceles triangle where the vertices A and B are given, and we know the angle at A and the length of side BC.

### Step-by-Step Solution:

1. **Identify the Coordinates of Points A and B:**
- We have point A at (-1, 0).
- Let point B be at (x_B, 0) since it lies on the x-axis. We will find the exact coordinates of B later.

2. **Determine the Length of AB:**
- Since AB = AC, we denote the length of AB as 'a'. We will find 'a' later.

3. **Angle at A:**
- Given that angle A = 120°, we can determine the angles at B and C. Since the triangle is isosceles, angles B and C will each be (180° - 120°) / 2 = 30°.

4. **Length of Side BC:**
- The length of BC is given as 4√3.

5. **Using Trigonometry to Find Coordinates of C:**
- We can drop a perpendicular from point C to the x-axis, creating a right triangle. Let the foot of the perpendicular be point C'.
- The angle at A is 120°, so the angle at C' is 30° (since the total angle at A is split into two angles of 30° each).

6. **Using the Length of BC:**
- The length of BC can be broken down into horizontal and vertical components using trigonometric ratios:
- Horizontal component (C'B) = BC * cos(30°) = 4√3 * (√3/2) = 6.
- Vertical component (C'C) = BC * sin(30°) = 4√3 * (1/2) = 2√3.

7. **Finding Coordinates of C:**
- Since B is at (x_B, 0), and we know that the horizontal distance from B to C' is 6, we can find the coordinates of C:
- If B is at (3, 0) (as we will find later), then C' will be at (3 + 6, 0) = (9, 0).
- The coordinates of C will then be (9, 2√3).

8. **Finding the Slope of BC:**
- The slope (m) of line BC can be calculated using the coordinates of B (3, 0) and C (9, 2√3):
- m = (y_C - y_B) / (x_C - x_B) = (2√3 - 0) / (9 - 3) = 2√3 / 6 = √3 / 3.

9. **Equation of Line BC:**
- Using the point-slope form of the line equation:
- y - y1 = m(x - x1)
- Using point B (3, 0) and slope m = √3/3:
- y - 0 = (√3/3)(x - 3)
- y = (√3/3)x - √3.

10. **Rearranging the Equation:**
- To express it in standard form:
- Multiply through by 3 to eliminate the fraction:
- 3y = √3x - 3√3
- Rearranging gives:
- √3x - 3y - 3√3 = 0.

11. **Final Equation:**
- The equation of line BC can be expressed as:
- x + √3y = 3.