The line 3x-3y+17=0 bisects the angle between a pair of lines of which one line is 2x+y+4=0, then the equation to the other line is

Solution:

$\begin{array}{l}Givenlines3x-3y+17=0--\left(1\right)\Rightarrow slope=1\\ 2x+y+4=0--\left(2\right)\Rightarrow slope=-2\\ solving\left(1\right)\&\left(2\right)\\ x y 1\\ \begin{array}{cccc}-3& 17& 3& -3\\ 1& 4& 2& 1\end{array}\\ \Rightarrow \frac{x}{-12-17}=\frac{y}{34-12}=\frac{1}{3+6}\\ \Rightarrow \frac{x}{-29}=\frac{y}{22}=\frac{1}{9}\\ \therefore P(\frac{-29}{9},\frac{22}{9})\end{array}$

$\begin{array}{l}Let\mathrm{Re}quiredEquationy-\frac{22}{9}=m(x+\frac{29}{9})--\left(3\right)\\ If\left(1\right)isanglebi\sec torof\left(2\right)\&\left(3\right)then\\ Angleb/w\left(1\right)\&\left(3\right)=Angleb/w\left(1\right)\&\left(2\right)\\ \therefore \left|\frac{m-1}{1+m}\right|=\left|\frac{1+2}{1-2}\right|\\ \Rightarrow \left|\frac{m-1}{1+m}\right|=3\\ If\frac{m-1}{1+m}=3,If\frac{m-1}{1+m}=-3\\ \Rightarrow m-1=3+3m,\Rightarrow m-1=-3-3m\\ \Rightarrow 2m=-4,\Rightarrow 4m=-2\\ \Rightarrow m=-2,\Rightarrow m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}$

$\begin{array}{l}\therefore m=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.from\left(3\right)\\ y-\frac{22}{9}=\frac{-1}{2}(x+\frac{29}{9})\\ \Rightarrow 2(9y-22)=-(9x+29)\\ \Rightarrow 18y-44=-9x-29\\ \Rightarrow 9x+18y-15=0\\ \Rightarrow 3x+6y-5=0\end{array}$