The lines `(lx+my)^2-3(mx-ly)^2=0` and `lx+my+n=0` forms

Options

A. an isosecles triangle
B. a right angled triangle
C. an equilateral triangle
D. None of these

Answer

To solve the problem, we need to analyze the given equations of the lines and find the angle between them.

### Step 1: Simplify the first equation
The first equation is given as:
\[ (lx + my)^2 - 3(mx - ly)^2 = 0 \]
We can expand both squares:
\[ (lx + my)^2 = l^2x^2 + m^2y^2 + 2lmyx \]
\[ (mx - ly)^2 = m^2x^2 - 2mlyx + l^2y^2 \]
Thus, we have:
\[ (lx + my)^2 - 3(mx - ly)^2 = l^2x^2 + m^2y^2 + 2lmyx - 3(m^2x^2 - 2mlyx + l^2y^2) = 0 \]

### Step 2: Combine like terms
Now, substituting the expansions back into the equation:
\[ l^2x^2 + m^2y^2 + 2lmyx - 3m^2x^2 + 6mlyx - 3l^2y^2 = 0 \]
Combining like terms gives:
\[ (l^2 - 3m^2)x^2 + (m^2 - 3l^2)y^2 + (2lm + 6ml)xy = 0 \]
This simplifies to:
\[ (l^2 - 3m^2)x^2 + (m^2 - 3l^2)y^2 + 8lmyx = 0 \]

### Step 3: Identify coefficients
From the combined equation, we can identify:
- \( a = l^2 - 3m^2 \)
- \( b = m^2 - 3l^2 \)
- \( h = 4lm \)

### Step 4: Use the formula for the angle between two lines
The angle \( \theta \) between the two lines can be found using the formula:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]

### Step 5: Substitute the values of \( a \), \( b \), and \( h \)
Substituting the values into the formula:
\[ \tan \theta = \frac{2\sqrt{(4lm)^2 - (l^2 - 3m^2)(m^2 - 3l^2)}}{(l^2 - 3m^2) + (m^2 - 3l^2)} \]

### Step 6: Simplify the expression
Calculating \( h^2 \):
\[ (4lm)^2 = 16l^2m^2 \]
Calculating \( ab \):
\[ ab = (l^2 - 3m^2)(m^2 - 3l^2) = l^2m^2 - 3l^4 - 3m^4 + 9m^2l^2 = 10l^2m^2 - 3l^4 - 3m^4 \]
Thus,
\[ h^2 - ab = 16l^2m^2 - (10l^2m^2 - 3l^4 - 3m^4) = 6l^2m^2 + 3l^4 + 3m^4 \]

### Step 7: Substitute back into the formula
Now substituting back into the formula for \( \tan \theta \):
\[ \tan \theta = \frac{2\sqrt{6l^2m^2 + 3l^4 + 3m^4}}{(l^2 - 3m^2) + (m^2 - 3l^2)} = \frac{2\sqrt{6l^2m^2 + 3l^4 + 3m^4}}{-2l^2 - 2m^2} \]
This simplifies to:
\[ \tan \theta = -\sqrt{\frac{6l^2m^2 + 3l^4 + 3m^4}{l^2 + m^2}} \]

### Step 8: Determine the angle
From the calculations, we find that:
\[ \tan \theta = \sqrt{3} \]
This implies:
\[ \theta = \frac{\pi}{3} \text{ or } 60^\circ \]

### Final Answer
The angle between the two lines is \( \frac{\pi}{3} \) radians or \( 60^\circ \).