The angle between the regression lines: `x-2y+3=0, 4x-5y+1=0` is

Options

A. `tan^(-1)(pm3/14)`
B. `pi/2`
C. `tan^(-1)(pm14/3)`
D. None of these

Answer

To find the angle between the regression lines given by the equations \(x - 2y + 3 = 0\) and \(4x - 5y + 1 = 0\), we will follow these steps:

### Step 1: Convert the equations into slope-intercept form
1. For the first equation \(x - 2y + 3 = 0\):
\[ 2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2} \]
Here, the slope \(m_1 = \frac{1}{2}\).

2. For the second equation \(4x - 5y + 1 = 0\):
\[ 5y = 4x + 1 \implies y = \frac{4}{5}x + \frac{1}{5} \]
Here, the slope \(m_2 = \frac{4}{5}\).

### Step 2: Use the formula for the tangent of the angle between two lines
The formula for the tangent of the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the values of \(m_1\) and \(m_2\)
Substituting \(m_1 = \frac{1}{2}\) and \(m_2 = \frac{4}{5}\) into the formula:
\[ \tan \theta = \left| \frac{\frac{4}{5} - \frac{1}{2}}{1 + \left(\frac{1}{2} \cdot \frac{4}{5}\right)} \right| \]

### Step 4: Simplify the expression
1. Calculate \(m_2 - m_1\):
\[ m_2 - m_1 = \frac{4}{5} - \frac{1}{2} = \frac{8}{10} - \frac{5}{10} = \frac{3}{10} \]

2. Calculate \(1 + m_1 m_2\):
\[ m_1 m_2 = \frac{1}{2} \cdot \frac{4}{5} = \frac{4}{10} = \frac{2}{5} \]
Therefore,
\[ 1 + m_1 m_2 = 1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5} \]

3. Substitute these values back into the tangent formula:
\[ \tan \theta = \left| \frac{\frac{3}{10}}{\frac{7}{5}} \right| = \left| \frac{3}{10} \cdot \frac{5}{7} \right| = \frac{15}{70} = \frac{3}{14} \]

### Step 5: Find \(\theta\)
Thus, we have:
\[ \tan \theta = \frac{3}{14} \]
To find \(\theta\), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{3}{14}\right) \]

### Final Answer
The angle between the regression lines is:
\[ \theta = \tan^{-1}\left(\frac{3}{14}\right) \]