Find the angle between the lines whose joint equation is `2x^2-3xy+y^2=0`

Answer

To find the angle between the lines given by the joint equation \(2x^2 - 3xy + y^2 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The joint equation of two lines can be compared to the standard form:
\[ ax^2 + 2hxy + by^2 = 0 \]
From the equation \(2x^2 - 3xy + y^2 = 0\), we can identify:
- \(a = 2\)
- \(h = -\frac{3}{2}\) (since the coefficient of \(xy\) is \(-3\), we take half of that)
- \(b = 1\)

### Step 2: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \(\theta\) between the two lines is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]

### Step 3: Substitute the values into the formula
Now, substituting the values of \(a\), \(b\), and \(h\) into the formula:
\[ \tan \theta = \frac{2\sqrt{(-\frac{3}{2})^2 - (2)(1)}}{2 + 1} \]

### Step 4: Calculate \(h^2 - ab\)
First, calculate \(h^2\) and \(ab\):
\[ h^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \]
\[ ab = 2 \cdot 1 = 2 \]
Now, substitute these into the formula:
\[ \tan \theta = \frac{2\sqrt{\frac{9}{4} - 2}}{3} \]
Convert \(2\) into a fraction with a denominator of \(4\):
\[ 2 = \frac{8}{4} \]
Now, we have:
\[ \tan \theta = \frac{2\sqrt{\frac{9}{4} - \frac{8}{4}}}{3} = \frac{2\sqrt{\frac{1}{4}}}{3} \]

### Step 5: Simplify the expression
\[ \tan \theta = \frac{2 \cdot \frac{1}{2}}{3} = \frac{1}{3} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \]

### Final Answer
Therefore, the angle between the lines is:
\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \]

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