The straight line $x+y+1=0$ bisects an angle between the pair of lines of which one is $2 x+3 y-4=0$. Then, the equation of the other line is

Let slope of required line $=m$ The angle between bisector and new line will be same as between bisector $x+y+1$ and line $2 x+3 y-4=0$ $\therefore$ Angle between bisector $x+y+1=0$ and new line is $$ \tan \theta=\left|\frac{m+1}{1-m}\right| $$ $\left[\because\right.$ Angle between two lines $\left.\tan \theta=\left|\frac{m_2-m_1}{\mid 1+m_1 m_2}\right|\right]$ And angle between bisector $x+y+1$ and $2 x+3 y-4=0$ $$ \begin{aligned} & \tan \theta=\left|\frac{\frac{-2}{3}+\frac{1}{3}}{1+\frac{2}{3} \times 1}\right|=\left|\frac{\frac{1}{3}}{\frac{3+2}{3}}\right| \\ & \tan \theta=\left|\frac{1}{5}\right| \end{aligned} $$ From Eqs. (i) and (ii), $$ \begin{aligned} & \frac{m+1}{1-m}=\left(\frac{1}{5}\right) \text { and } \frac{m+1}{1-m}=\frac{-1}{5} \\ & \Rightarrow 5 m+5=1-m \text { or } 5 m+5=-1+m \\ & \Rightarrow 6 m=-4 \text { or } 4 m=-6 \end{aligned} $$ $\Rightarrow m=\frac{-2}{3}$ or $m=\frac{-3}{2}$ So, new line has slope of $\frac{-3}{2}$. Now, for point of intersection of line and bisector, we solve, $$ \text { and } \quad \begin{aligned} x+y & =-1 \\ \Rightarrow \quad 2 x+3 y & =4 \\ 2 x+2 y & =-2 \\ 2 x \pm 3 y & =4 \\ \hline-y & =-6 \\ y & =6 \end{aligned} $$ Put in Eq. (iii) $x+y=-1$ $$ \Rightarrow \quad x+6=-1 \Rightarrow x=-7 $$ So, coordinate of point of intersection is $(-7,6)$ equation of line passes through $(-7,6)$ and slope $\frac{-3}{2}$ $$ \begin{aligned} y-y_1=m\left(x-x_1\right) & \Rightarrow(y-6)=\frac{-3}{2}(x+7) \\ \Rightarrow 2 y-12=-3 x-21 & \Rightarrow 3 x+2 y+9=0 \end{aligned} $$