Find the bisector of the angle in which the origin lies, between the pair of straight lines $\;$ $y - b = \dfrac{2m}{1 - m^2} \left(x - a\right)$ $\;$ and $\;$ $y - b = \dfrac{2m_1}{1 - m_1^2} \left(x - a\right)$.
Equations of the given lines are:
$y - b = \dfrac{2m}{1 - m^2} \left(x - a\right)$ $\;\;\; \cdots \; (1)$
$y - b = \dfrac{2m_1}{1 - m_1^2} \left(x - a\right)$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be rewritten as
$\left(1 - m^2\right) \left(y - b\right) = 2m \left(x - a\right)$
i.e. $\;$ $\left(1 - m^2\right)y - b + m^2 b = 2mx - 2ma$
i.e. $\;$ $2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right) = 0$ $\;\;\; \cdots \; (1a)$
Equation $(2)$ can be rewritten as
$\left(1 - m_1^2\right) \left(y - b\right) = 2m_1 \left(x - a\right)$
i.e. $\;$ $\left(1 - m_1^2\right)y - b + m_1^2 b = 2m_1x - 2m_1a$
i.e. $\;$ $2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right) = 0$ $\;\;\; \cdots \; (2a)$
$\therefore \;$ The equation to the bisector of the angle in which the origin lies is
$\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{\left(2m\right)^2 + \left(1 - m^2\right)^2}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{\left(2m_1\right)^2 + \left(1 - m_1^2\right)^2}}$
i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{4m^2 + 1 - 2m^2 + m^4}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{4m_1^2 + 1 - 2m_1^2 + m_1^4}}$
i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{2m^2 + 1 + m^4}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{2m_1^2 + 1 + m_1^4}}$
i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{\left(1 + m^2\right)^2}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{\left(1 + m_1^2\right)^2}}$
i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{1 + m^2}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{1 + m_1^2}$
i.e. $\;$ $2m \left(1 + m_1^2\right) x - \left(1 - m^2\right) \left(1 + m_1^2\right)y - \left(2ma - b + m^2b\right) \left(1 + m_1^2\right)$
$\hspace{1cm}$ $= 2m_1 \left(1 + m^2\right) x - \left(1 - m_1^2\right) \left(1 + m^2\right)y - \left(2m_1a - b + m_1^2b\right) \left(1 + m^2\right)$
i.e. $\;$ $2\left(m + m m_1^2 - m_1 - m_1 m^2\right)x - \left(1 + m_1^2 - m^2 - m^2 m_1^2 -1 - m^2 + m_1^2 + m^2 m_1^2\right)y$
$\hspace{1cm}$ $= 2ma + 2mm_1^2 a - b - b m_1^2 + bm^2 + bm^2 m_1^2$
$\hspace{1.5cm}$ $-2m_1a - 2m_1 m^2 a + b + bm^2 - b m_1^2 - b m^2 m_1^2$
i.e. $\;$ $x \left(m + m m_1^2 - m_1 - m_1 m^2\right) - y \left(m_1^2 - m^2\right)$
$\hspace{1cm}$ $= ma + m m_1^2 a - b m_1^2 + bm^2 - m_1 a - m_1 m^2 a$
i.e. $\;$ $x \left(m + m m_1^2 - m_1 - m_1 m^2\right) - y \left(m_1^2 - m^2\right)$
$\hspace{1cm}$ $= a \left(m + m m_1^2 - m_1 - m_1 m^2\right) - b \left(m_1^2 - m^2\right)$
i.e. $\;$ $\left(x - a\right) \left(m + m m_1^2 - m_1 - m_1 m^2\right) - \left(y - b\right) \left(m_1^2 - m^2\right) = 0$
i.e. $\;$ $\left(x - a\right) \left[\left(m - m_1\right) + mm_1 \left(m_1 - m\right)\right] - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$
i.e. $\;$ $\left(x - a\right) \left[- \left(m_1 - m\right) + mm_1 \left(m_1 - m\right)\right] - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$
i.e. $\;$ $\left(x - a\right) \left(mm_1 - 1\right) \left(m_1 - m\right) - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$
i.e. $\;$ $\left(x - a\right) \left(mm_1 - 1\right) - \left(y - b\right) \left(m_1 + m\right) = 0$