If the lines y = 3x + 1 and 2y = x + 3 are equally

Solution:

The equations of the given lines are

y = 3x + 1 … (1)

2y = x + 3 … (2)

y = mx + 4 … (3)

Slope of line $(1), m_{1}=3$

Slope of line $(2), m_{2}=\frac{1}{2}$

Slope of line $(3), m_{3}=m$

It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).

$\therefore\left|\frac{m_{1}-m_{3}}{1+m_{1} m_{3}}\right|=\left|\frac{m_{2}-m_{3}}{1+m_{2} m_{3}}\right|$

$\Rightarrow\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{1}{2} m}\right|$

$\Rightarrow\left|\frac{3-m}{1+3 m}\right|=\left|\frac{1-2 m}{m+2}\right|$

$\Rightarrow \frac{3-m}{1+3 m}=\pm\left(\frac{1-2 m}{m+2}\right)$

$\Rightarrow \frac{3-m}{1+3 m}=\frac{1-2 m}{m+2}$ or $\frac{3-m}{1+3 m}=-\left(\frac{1-2 m}{m+2}\right)$

If $\frac{3-m}{1+3 m}=\frac{1-2 m}{m+2}$, then'

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

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

$\Rightarrow 5 m^{2}+5=0$

$\Rightarrow\left(m^{2}+1\right)=0$

$\Rightarrow m=\sqrt{-1}$, which is not real

Hence, this case is not posible.

If $\frac{3-m}{1+3 m}=-\left(\frac{1-2 m}{m+2}\right)$, then

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

$\Rightarrow-m^{2}+m+6=-\left(1+m-6 m^{2}\right)$

$\Rightarrow 7 m^{2}-2 m-7=0$

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

$\Rightarrow m=\frac{2 \pm 2 \sqrt{1+49}}{14}$

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

Thus, the required value of $m$ is $\frac{1 \pm 5 \sqrt{2}}{7}$.