Question

The equation of the bisector of the angle between two lines $3x-4y+12=0$ and $12x-5y+7=0,$ which contain the point $(-1,4)$ is

Answer 1

The equation of the bisector of the angle between two lines consisting of the point $(\alpha ,\beta )$ is
$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{{a_1}^2+{b_1}^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{{a_2}^2+{b_2}^2}}$
The equation is $\left[\frac{(3x–4y+12)}{5}\right]=\pm \left[\frac{(12x–5y+7)}{13}\right]$ at point $(-1,4)$ value of $\left[\frac{(3x–4y+12)}{(12x–5y+7)}\right]>0$
On taking the positive sign,
$$\begin{gathered} \left[\frac{(3x–4y+12)}{5}\right]=\left[\frac{(12x–5y+7)}{13}\right] \\ 39x–52y+156=60x–25y+35 \\ 21x+27y–121=0 \end{gathered}$$

Hence, the required equation of the bisector of the angle between two lines is $12x+27y-121=0$