Show that the straight lines `x^2+4xy+y^2=0 ` and the line x-y=4 form an equilateral triangle .

Answer

To show that the straight lines \(x^2 + 4xy + y^2 = 0\) and the line \(x - y = 4\) form an equilateral triangle, we will follow these steps:

### Step 1: Find the equations of the lines from the quadratic equation

The equation \(x^2 + 4xy + y^2 = 0\) can be factored or solved to find the slopes of the lines it represents. We can rewrite it in terms of \(y\):

\[ y^2 + 4xy + x^2 = 0 \]

This is a quadratic equation in \(y\). Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = 4x\), and \(c = x^2\).

Now, calculate the discriminant:

\[ D = (4x)^2 - 4(1)(x^2) = 16x^2 - 4x^2 = 12x^2 \]

Now, substituting back into the quadratic formula:

\[ y = \frac{-4x \pm \sqrt{12x^2}}{2} = \frac{-4x \pm 2\sqrt{3}x}{2} = -2x \pm \sqrt{3}x \]

Thus, we have two lines:

1. \(y = (-2 + \sqrt{3})x\)
2. \(y = (-2 - \sqrt{3})x\)

### Step 2: Find the slope of the line \(x - y = 4\)

Rearranging \(x - y = 4\) gives us:

\[ y = x - 4 \]

The slope of this line is \(m_3 = 1\).

### Step 3: Calculate the angles between the lines

Now we have three slopes:

1. \(m_1 = -2 + \sqrt{3}\)
2. \(m_2 = -2 - \sqrt{3}\)
3. \(m_3 = 1\)

To find the angles between these lines, we will use the formula for the tangent of the angle between two lines:

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Calculating \(m_1 - m_2\):

\[ m_1 - m_2 = (-2 + \sqrt{3}) - (-2 - \sqrt{3}) = 2\sqrt{3} \]

Calculating \(1 + m_1 m_2\):

\[ m_1 m_2 = (-2 + \sqrt{3})(-2 - \sqrt{3}) = 4 - 3 = 1 \]

Thus,

\[ 1 + m_1 m_2 = 1 + 1 = 2 \]

Now substituting into the tangent formula:

\[ \tan \theta = \left| \frac{2\sqrt{3}}{2} \right| = \sqrt{3} \]

This implies:

\[ \theta = 60^\circ \]

### Step 4: Check the angle between the other pairs of lines

Now we need to check the angle between the lines \(m_1\) and \(m_3\) and between \(m_2\) and \(m_3\).

Using the same formula for \(m_1\) and \(m_3\):

\[ \tan \theta = \left| \frac{(-2 + \sqrt{3}) - 1}{1 + (-2 + \sqrt{3}) \cdot 1} \right| \]

Calculating:

\[ -2 + \sqrt{3} - 1 = -3 + \sqrt{3} \]
\[ 1 + (-2 + \sqrt{3}) = -1 + \sqrt{3} \]

Thus,

\[ \tan \theta = \left| \frac{-3 + \sqrt{3}}{-1 + \sqrt{3}} \right| \]

This calculation will also yield \(60^\circ\).

Repeating for \(m_2\) and \(m_3\) will yield the same result.

### Conclusion

Since all angles between the lines are \(60^\circ\), the lines \(x^2 + 4xy + y^2 = 0\) and \(x - y = 4\) form an equilateral triangle.