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 solve the problem of finding the equations of the other two sides of the equilateral triangle with 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: Identify the Slope of the Given Line
The equation of the line can be rewritten as:
\[ y = -x + 1 \]
From this, we can see that the slope (m) of the line is -1.

### Step 3: Find the Slope of the Other Two Sides
In an equilateral triangle, the angles between the sides are 60 degrees. Therefore, we can find the slopes of the other two sides using the formula for the tangent of the angle between two lines:
\[ \tan(60^\circ) = \sqrt{3} \]
Let the slopes of the other two sides be \( m_1 \) and \( m_2 \). We know:
- The slope of side BC (along the line) is \( m_{BC} = -1 \).

Using the formula:
\[ \tan(60^\circ) = \left| \frac{m_{BC} - m_1}{1 + m_{BC} m_1} \right| \]
Substituting \( m_{BC} = -1 \):
\[ \sqrt{3} = \left| \frac{-1 - m_1}{1 - m_1} \right| \]

### Step 4: Solve for the Slopes
This gives us two cases to solve:

**Case 1:**
\[ \sqrt{3} = \frac{-1 - m_1}{1 - m_1} \]
Cross-multiplying and simplifying leads to:
\[ \sqrt{3}(1 - m_1) = -1 - m_1 \]
\[ \sqrt{3} - \sqrt{3}m_1 = -1 - m_1 \]
Rearranging gives:
\[ m_1(\sqrt{3} - 1) = \sqrt{3} + 1 \]
Thus:
\[ m_1 = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

**Case 2:**
\[ \sqrt{3} = \frac{1 + m_1}{1 + m_1} \]
This leads to:
\[ \sqrt{3}(1 + m_1) = -1 - m_1 \]
Rearranging gives:
\[ m_1(\sqrt{3} + 1) = -1 - \sqrt{3} \]
Thus:
\[ m_1 = \frac{-1 - \sqrt{3}}{\sqrt{3} + 1} \]

### Step 5: Find the Equations of the Lines
Now we have two slopes \( m_1 \) and \( m_2 \). We can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Using vertex A (-1, -1) for both slopes.

For the first slope \( m_1 = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \):
\[ y + 1 = m_1(x + 1) \]

For the second slope \( m_2 = \frac{-1 - \sqrt{3}}{\sqrt{3} + 1} \):
\[ y + 1 = m_2(x + 1) \]

### Final Equations
1. For \( m_1 \):
\[ y + 1 = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}(x + 1) \]

2. For \( m_2 \):
\[ y + 1 = \frac{-1 - \sqrt{3}}{\sqrt{3} + 1}(x + 1) \]