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

Options

A. an isosceles triangle
B. a right angled triangle
C. an equilateral triangle
D. None of the above

Explanation

To solve the problem, we need to analyze the given equations and find the angles formed by the lines represented by them.

### Step 1: Expand the first equation
The first equation is given as:
\[ (lx + my)^2 - 3(mx - ly)^2 = 0 \]
We will expand both squares:
\[ (lx + my)^2 = l^2x^2 + m^2y^2 + 2lmyx \]
\[ (mx - ly)^2 = m^2x^2 + l^2y^2 - 2lmyx \]
Now substituting these into the equation:
\[ l^2x^2 + m^2y^2 + 2lmyx - 3(m^2x^2 + l^2y^2 - 2lmyx) = 0 \]
Expanding this gives:
\[ l^2x^2 + m^2y^2 + 2lmyx - 3m^2x^2 - 3l^2y^2 + 6lmyx = 0 \]
Combining like terms results in:
\[ (l^2 - 3m^2)x^2 + (m^2 - 3l^2)y^2 + 8lmyx = 0 \]

### Step 2: Identify coefficients
From the combined equation, we can identify:
- \( A = l^2 - 3m^2 \)
- \( B = 8lm \)
- \( C = m^2 - 3l^2 \)

### Step 3: Calculate the angle between the lines
The formula for the angle \( \theta \) between two lines represented by the general equation \( Ax^2 + Bxy + Cy^2 = 0 \) is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
where \( h = \frac{B}{2} \), \( a = A \), and \( b = C \).

Substituting the values we found:
- \( h = \frac{8lm}{2} = 4lm \)
- \( a = l^2 - 3m^2 \)
- \( b = m^2 - 3l^2 \)

Thus,
\[ \tan \theta = \frac{2 \cdot 4lm}{(l^2 - 3m^2) + (m^2 - 3l^2)} \]

### Step 4: Simplify the expression
Calculating the denominator:
\[ l^2 - 3m^2 + m^2 - 3l^2 = -2l^2 - 2m^2 = -2(l^2 + m^2) \]
So we have:
\[ \tan \theta = \frac{8lm}{-2(l^2 + m^2)} = -\frac{4lm}{l^2 + m^2} \]

### Step 5: Find the angle
The value of \( \tan \theta \) can be simplified further. If we assume \( l \) and \( m \) are positive, we can consider the absolute value:
\[ |\tan \theta| = \frac{4lm}{l^2 + m^2} \]
To find the specific angle, we can analyze the situation further. If \( \tan \theta = \sqrt{3} \), then:
\[ \theta = 60^\circ \text{ or } \frac{\pi}{3} \]

### Step 6: Determine the triangle type
Since we have two angles of \( 60^\circ \), the third angle will be:
\[ 180^\circ - 60^\circ - 60^\circ = 60^\circ \]
Thus, all angles are \( 60^\circ \), indicating that the triangle formed is an equilateral triangle.

### Final Answer
The lines form an **equilateral triangle**.

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