The angle between the pair of lines represented by `2x^2-7xy+3y^2=0` is

Options

A. `60^@`
B. `45^@`
C. `tan^(-1).((7)/(6))`
D. `30^@`

Explanation

To find the angle between the pair of lines represented by the equation \(2x^2 - 7xy + 3y^2 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given equation is in the form \(Ax^2 + Bxy + Cy^2 = 0\), where:
- \(A = 2\)
- \(B = -7\)
- \(C = 3\)

### Step 2: Use the formula for the angle between two lines
The angle \(\theta\) between the two lines represented by the equation can be found using the formula:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
where \(m_1\) and \(m_2\) are the slopes of the lines.

### Step 3: Calculate the slopes using the quadratic formula
The slopes can be found using the formula:
\[ m = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values:
\[ m = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \]
\[ = \frac{7 \pm \sqrt{49 - 24}}{4} \]
\[ = \frac{7 \pm \sqrt{25}}{4} \]
\[ = \frac{7 \pm 5}{4} \]
This gives us two slopes:
\[ m_1 = \frac{12}{4} = 3 \quad \text{and} \quad m_2 = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Substitute the slopes into the angle formula
Now, substitute \(m_1\) and \(m_2\) into the angle formula:
\[ \tan \theta = \frac{3 - \frac{1}{2}}{1 + 3 \cdot \frac{1}{2}} \]
\[ = \frac{\frac{6}{2} - \frac{1}{2}}{1 + \frac{3}{2}} \]
\[ = \frac{\frac{5}{2}}{\frac{5}{2}} = 1 \]

### Step 5: Find the angle
Since \(\tan \theta = 1\), we have:
\[ \theta = \tan^{-1}(1) = \frac{\pi}{4} \text{ radians} \text{ or } 45^\circ \]

### Final Answer
The angle between the pair of lines represented by \(2x^2 - 7xy + 3y^2 = 0\) is \(\frac{\pi}{4}\) radians or \(45^\circ\).

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