An equilateral triangle has its circumcentre at origin and one of the sides is along the line x + y -1 = 0 and cordinate of vertex A is (-1,-1). Find the equations of the other two sides.

Answer

To find the equations of the other two sides of the equilateral triangle with the given conditions, we can follow these steps:

### Step 1: Understand the Given Information
We have:
- An equilateral triangle with circumcenter at the origin (0, 0).
- One vertex A at (-1, -1).
- One side along the line \(x + y - 1 = 0\).

### Step 2: Determine the Slope of the Given Line
The equation of the line can be rearranged to slope-intercept form:
\[ y = -x + 1 \]
The slope (m1) of this line is -1.

### Step 3: Calculate the Slopes of the Other Two Sides
Since the triangle is equilateral, the angles between the sides and the line must be 60 degrees. We can use the formula for the tangent of the angle between two lines:
\[ \tan(\theta) = \frac{|m_1 - m_2|}{1 + m_1 m_2} \]
For \( \theta = 60^\circ \), we have \( \tan(60^\circ) = \sqrt{3} \).

Setting up the equation:
\[ \sqrt{3} = \frac{|-1 - m_2|}{1 - m_2} \]

### Step 4: Solve for m2
1. **Case 1:** \( -1 - m_2 = \sqrt{3}(1 - m_2) \)
\[ -1 - m_2 = \sqrt{3} - \sqrt{3}m_2 \]
Rearranging gives:
\[ m_2(\sqrt{3} - 1) = \sqrt{3} + 1 \implies m_2 = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]
Rationalizing gives:
\[ m_2 = \frac{(\sqrt{3} + 1)^2}{2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

2. **Case 2:** \( -1 - m_2 = -\sqrt{3}(1 - m_2) \)
\[ -1 - m_2 = -\sqrt{3} + \sqrt{3}m_2 \]
Rearranging gives:
\[ m_2(1 + \sqrt{3}) = -1 + \sqrt{3} \implies m_2 = \frac{-1 + \sqrt{3}}{1 + \sqrt{3}} \]
Rationalizing gives:
\[ m_2 = \frac{(-1 + \sqrt{3})(1 - \sqrt{3})}{-2} = \frac{-1 + \sqrt{3} + \sqrt{3} - 3}{-2} = \frac{2\sqrt{3} - 4}{-2} = 2 - \sqrt{3} \]

### Step 5: Write the Equations of the Lines
Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\):

1. For the slope \(2 + \sqrt{3}\):
\[ y + 1 = (2 + \sqrt{3})(x + 1) \]
Rearranging gives:
\[ y = (2 + \sqrt{3})x + (1 + 2 + \sqrt{3}) - 1 = (2 + \sqrt{3})x + 2 + \sqrt{3} \]

2. For the slope \(2 - \sqrt{3}\):
\[ y + 1 = (2 - \sqrt{3})(x + 1) \]
Rearranging gives:
\[ y = (2 - \sqrt{3})x + (1 + 2 - \sqrt{3}) - 1 = (2 - \sqrt{3})x + 2 - \sqrt{3} \]

### Final Result
The equations of the other two sides of the triangle are:
1. \(y = (2 + \sqrt{3})x + 1 + \sqrt{3}\)
2. \(y = (2 - \sqrt{3})x + 1 - \sqrt{3}\)