If the lines `y=x+3 and y= 3x+1` are equally inclined to the line `y= mx+4` then the value of m is

Options

A. `(1+ 5sqrt2)/(2)`
B. `(5 sqrt2-1)/(7)`
C. `(1- sqrt5)/(2)`
D. None of these

Answer

To solve the problem, we need to find the value of \( m \) such that the lines \( y = x + 3 \) and \( y = 3x + 1 \) are equally inclined to the line \( y = mx + 4 \).

### Step 1: Identify the slopes of the given lines
The slope of the line \( y = x + 3 \) is:
\[ m_1 = 1 \]
The slope of the line \( y = 3x + 1 \) is:
\[ m_2 = 3 \]
The slope of the line \( y = mx + 4 \) is:
\[ m = m \]

### Step 2: Use the formula for the tangent of the angle between two lines
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
We need to find the angles between the line \( y = mx + 4 \) and the other two lines.

### Step 3: Set up the equations for the angles
For the line \( y = x + 3 \) (slope \( m_1 = 1 \)):
\[ \tan \theta_1 = \left| \frac{m - 1}{1 + m \cdot 1} \right| = \left| \frac{m - 1}{1 + m} \right| \]

For the line \( y = 3x + 1 \) (slope \( m_2 = 3 \)):
\[ \tan \theta_2 = \left| \frac{m - 3}{1 + m \cdot 3} \right| = \left| \frac{m - 3}{1 + 3m} \right| \]

### Step 4: Set the two angles equal since they are equally inclined
Since the lines are equally inclined to \( y = mx + 4 \):
\[ \left| \frac{m - 1}{1 + m} \right| = \left| \frac{m - 3}{1 + 3m} \right| \]

### Step 5: Solve the equation
This leads to two cases to consider:

#### Case 1: Both expressions are positive
\[ \frac{m - 1}{1 + m} = \frac{m - 3}{1 + 3m} \]
Cross-multiplying gives:
\[ (m - 1)(1 + 3m) = (m - 3)(1 + m) \]
Expanding both sides:
\[ m + 3m^2 - 1 - 3m = m + m^2 - 3 - 3m \]
Simplifying:
\[ 3m^2 - 2m - 1 = 0 \]

#### Case 2: One expression is positive and the other is negative
\[ \frac{m - 1}{1 + m} = -\frac{m - 3}{1 + 3m} \]
Cross-multiplying gives:
\[ (m - 1)(1 + 3m) = -(m - 3)(1 + m) \]
Expanding both sides:
\[ m + 3m^2 - 1 - 3m = -m - m^2 + 3 + 3m \]
Simplifying:
\[ 4m^2 + 4m - 4 = 0 \]

### Step 6: Solve the quadratic equations
1. From \( 3m^2 - 2m - 1 = 0 \):
\[ m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6} \]
This gives:
\[ m = 1 \quad \text{or} \quad m = -\frac{1}{3} \]

2. From \( 4m^2 + 4m - 4 = 0 \):
\[ m = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} = \frac{-4 \pm \sqrt{16 + 64}}{8} = \frac{-4 \pm 8}{8} \]
This gives:
\[ m = \frac{1}{2} \quad \text{or} \quad m = -\frac{3}{2} \]

### Final Answer
The possible values of \( m \) are \( 1, -\frac{1}{3}, \frac{1}{2}, -\frac{3}{2} \).