If the straight line $2 x+3 y+1=0$ bisects the angle between a pair of lines, one of which in this pair is…

1. $3 x+4 y-9=0$
2. 6x - 7 y - 14 = 0
3. 9x + 46y - 28 = 0
4. 9x - 23y - 12 = 0

Solution

Let the angle bisector of $ \begin{aligned} & 3 x+2 y+4 \text { and } \\ & y=m x+c \text { is } \\ & \quad 2 x+3 y+1=0 \end{aligned} $ Slope of line $ \begin{aligned} & 3 x+2 y+4 \text { is } \\ & \qquad m_1=\frac{-3}{2} \end{aligned} $
Slope of line $2 x+3 y+1=0$ is $m_2=\frac{-2}{3}$ Angle between lines and bisector are equal, $ \begin{aligned} & \text { so }\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{m-m_2}{1+m_1 m_2}\right| \\ & \Rightarrow \frac{-2 / 3+3 / 2}{1+1}=\frac{m+2 / 3}{1-\frac{2 m}{3}} \Rightarrow \frac{5}{12}=\frac{m+2 / 3}{1-2 m / 3} \\ & \Rightarrow \quad 5-\frac{10 m}{3}=12 m+8 \Rightarrow\left(\frac{10}{3}+12\right) m=-3 \\ & \Rightarrow \quad \frac{46}{3} m=-3 \Rightarrow m=-\frac{9}{46} \end{aligned} $ So, equation of required line is
Now, for intersection point, equation $3 x+2 y+4=0$ and $2 x+3 y+1=0$, is given by $ \begin{aligned} 6 x+4 y+8 & =0 \\ 6 x+9 y+3 & =0 \\ -\quad-\quad & \\ \hline 5 y & =5 \\ y & =1 \\ x & =-2 \end{aligned} $ and intersection point is $(-2,1)$. Required line is also passes through $(-2,1)$ $ \text { So, } 1=\frac{-9}{46}(-2)+c \Rightarrow c=1-\frac{9}{23} \Rightarrow c=\frac{14}{23} $ Put in Eq. (i), we get $ \begin{aligned} y & =\frac{-9}{46} x+\frac{14}{23} \\ \Rightarrow \quad 46 y & =-9 x+28 \Rightarrow 9 x+46 y-28=0 \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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