$\begin {array} {1 1} (A)\;\large\frac{-1 \pm 5 \sqrt 2}{7} & \quad (B)\;\large\frac{1 \pm 5 \sqrt 2}{7} \\ (C)\;\large\frac{-1 \pm 2\sqrt 5}{7} & \quad (D)\;\large\frac{1 \pm 2\sqrt 5}{7} \end {array}$

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- Angle between two lines whose slopes are $m_1$ and $m_2$ are $ \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$

The equation of the given lines are

$ \quad y=3x+1$-------(1)

$ \quad 2y=x+3$--------(2)

$ \quad y=mx+4$--------(3)

Let the slope of line (1) be $m_1=3$

Let the slope of line (2) be $m_2= \large\frac{1}{2}$

Let the slope of line (3) be $m_3=m$

Since it is given that the line (1) and (2) are equally inclined to line (3),

$ \bigg| \large\frac{m_1-m_3}{1+m_1m_3} \bigg|$$ = \bigg| \large\frac{m_2-m_3}{1+m_2m_3} \bigg|$

Substituting the values we get,

$\bigg| \large\frac{ 3-m}{1+3m} \bigg|$$ = \bigg| \large\frac{\Large\frac{1}{2}-m}{1+ \Large\frac{1}{2}m} \bigg|$

$ \Rightarrow \bigg| \large\frac{3-m}{1+3m} \bigg|$$ = \bigg| \large\frac{1-2m}{2+m} \bigg|$

i.e., $ \large\frac{3-m}{1+3m} = \pm \bigg( \large\frac{1-2m}{2+m} \bigg)$

Consider the case when

$ \large\frac{3-m}{1+3m} = \bigg( \large\frac{1-2m}{2+m} \bigg)$

On simplifying we get,

$ (3-m)(2+m)= (1-2m)(1+3m)$

$ \Rightarrow -m^2+m+6 = 1 + m -6m^2$

(i.e., ) $5m^2+5=0$

$ \Rightarrow m = \sqrt{-1}$

which is not real.

Hence it is not admissable.

Let us consider the case when

$ \large\frac{3-m}{1+3m}$$ = \large\frac{-(1-2m)}{m+2}$

$ \Rightarrow (3-m)(m+2) = -(1-2m)(1+3m)$

$ \Rightarrow -m^2+m+6 = -(1+m-6m^2)$

i.e., $7m^2-2m-7=0$

Let us apply the quadratic formula to solve the above equation.

$ \large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values for $a=7, b=-2$ and $c=-7$ we get,

$ m = \large\frac{-(-2) \pm \sqrt{4-4(-7)(7)}}{2 \times 7}$

$ = \large\frac{2 \pm \sqrt{4+196}}{14}$

$ = \large\frac{2 \pm \sqrt{200}}{14}$

$ = \large\frac{2 \pm 10\sqrt 2}{14}$

$ = \large\frac{1 \pm 5\sqrt 2}{7}$

Hence the value of $m$ is $ \large\frac{1 \pm 5\sqrt 2}{7}$

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