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

# 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

1. A

$3x+6y-5=0$

2. B

$3x+6y-7=0$

3. C

$7x-y+14=0$

4. D

$4x-y+3=0$

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### Solution:

$\begin{array}{l}Let\text{\hspace{0.17em}}\mathrm{Re}quired\text{\hspace{0.17em}}Equation\text{\hspace{0.17em}}y-\frac{22}{9}=m\left(x+\frac{29}{9}\right)--\left(3\right)\\ If\text{\hspace{0.17em}}\left(1\right)\text{\hspace{0.17em}\hspace{0.17em}}is\text{\hspace{0.17em}\hspace{0.17em}}angle\text{\hspace{0.17em}}bi\mathrm{sec}tor\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\left(2\right)&\left(3\right)\text{\hspace{0.17em}\hspace{0.17em}}then\\ Angle\text{\hspace{0.17em}}b/w\text{\hspace{0.17em}}\left(1\right)&\left(3\right)=\text{\hspace{0.17em}}Angle\text{\hspace{0.17em}}b/w\text{\hspace{0.17em}}\left(1\right)&\left(2\right)\\ \therefore \text{\hspace{0.17em}}|\frac{m-1}{1+m}|=|\frac{1+2}{1-2}|\\ ⇒\text{\hspace{0.17em}}|\frac{m-1}{1+m}|=3\\ If\text{\hspace{0.17em}}\frac{m-1}{1+m}=3,If\text{\hspace{0.17em}}\frac{m-1}{1+m}=-3\\ ⇒\text{\hspace{0.17em}}m-1=3+3m,⇒\text{\hspace{0.17em}}m-1=-3-3m\\ ⇒\text{\hspace{0.17em}}2m=-4,⇒\text{\hspace{0.17em}}4m=-2\\ ⇒\text{\hspace{0.17em}}m=-2,⇒\text{\hspace{0.17em}}m=-1}{2}\end{array}$

$\begin{array}{l}\therefore \text{\hspace{0.17em}}m=-1}{2}\text{\hspace{0.17em}\hspace{0.17em}}from\text{\hspace{0.17em}}\left(3\right)\\ y-\frac{22}{9}=\frac{-1}{2}\left(x+\frac{29}{9}\right)\\ ⇒\text{\hspace{0.17em}}2\left(9y-22\right)=-\left(9x+29\right)\\ ⇒\text{\hspace{0.17em}}18y-44=-9x-29\\ ⇒\text{\hspace{0.17em}}9x+18y-15=0\\ ⇒\text{\hspace{0.17em}}3x+6y-5=0\end{array}$

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