The equation `x^(3)-6x^(2)y+11xy^(2)-6y^(3)=0` represents three straight lines passing through the origin, the slopes of which form an

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the given cubic equation and determine the slopes of the lines it represents. The equation is:

\[ x^3 - 6x^2y + 11xy^2 - 6y^3 = 0 \]

### Step 1: Rewrite the Equation
We start by rewriting the equation in a more manageable form. We can divide the entire equation by \( x^3 \):

\[ 1 - \frac{6y}{x} + \frac{11y^2}{x^2} - \frac{6y^3}{x^3} = 0 \]

Let \( m = \frac{y}{x} \). Then, we can substitute \( \frac{y}{x} \) with \( m \):

\[ 1 - 6m + 11m^2 - 6m^3 = 0 \]

### Step 2: Factor the Cubic Equation
Now we have a cubic equation in terms of \( m \):

\[ -6m^3 + 11m^2 - 6m + 1 = 0 \]

To find the roots, we can check for rational roots. Testing \( m = 1 \):

\[ -6(1)^3 + 11(1)^2 - 6(1) + 1 = -6 + 11 - 6 + 1 = 0 \]

Thus, \( m = 1 \) is a root. We can factor out \( (m - 1) \).

### Step 3: Polynomial Long Division
Now we will divide the cubic polynomial by \( (m - 1) \):

Using synthetic division or polynomial long division, we divide:

\[ -6m^3 + 11m^2 - 6m + 1 \div (m - 1) \]

1. The first term is \( -6m^2 \) (multiply \( (m - 1) \) by \( -6m^2 \)).
2. Subtract to get \( 5m^2 - 6m + 1 \).
3. The next term is \( 5m \) (multiply \( (m - 1) \) by \( 5m \)).
4. Subtract to get \( -m + 1 \).
5. The last term is \( -1 \) (multiply \( (m - 1) \) by \( -1 \)).
6. Subtract to get \( 0 \).

Thus, we can express the cubic polynomial as:

\[ (m - 1)(-6m^2 + 5m - 1) = 0 \]

### Step 4: Solve the Quadratic Equation
Now we need to solve the quadratic equation:

\[ -6m^2 + 5m - 1 = 0 \]

Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = -6, b = 5, c = -1 \):

\[ m = \frac{-5 \pm \sqrt{(5)^2 - 4(-6)(-1)}}{2(-6)} \]
\[ = \frac{-5 \pm \sqrt{25 - 24}}{-12} \]
\[ = \frac{-5 \pm 1}{-12} \]

Calculating the two roots:

1. \( m_1 = \frac{-4}{-12} = \frac{1}{3} \)
2. \( m_2 = \frac{-6}{-12} = \frac{1}{2} \)

### Step 5: List the Slopes
So, the slopes of the three lines are:

1. \( m_1 = 1 \)
2. \( m_2 = \frac{1}{2} \)
3. \( m_3 = \frac{1}{3} \)

### Step 6: Check for Progressions
Now we need to check if these slopes form an Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

1. **AP Check**: For three numbers \( a, b, c \) to be in AP, \( 2b = a + c \).
- Here, \( 2 \cdot \frac{1}{2} = 1 + \frac{1}{3} \) → \( 1 = \frac{4}{3} \) (not true).

2. **GP Check**: For three numbers \( a, b, c \) to be in GP, \( b^2 = ac \).
- Here, \( \left(\frac{1}{2}\right)^2 = 1 \cdot \frac{1}{3} \) → \( \frac{1}{4} = \frac{1}{3} \) (not true).

3. **HP Check**: For three numbers to be in HP, their reciprocals must be in AP.
- The reciprocals are \( 1, 2, 3 \) which are in AP.

### Conclusion
Thus, the slopes \( 1, \frac{1}{2}, \frac{1}{3} \) form a Harmonic Progression (HP).

### Final Answer
The slopes of the lines represented by the equation form a Harmonic Progression (HP).

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