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

I know how to take out the equation of the 2 bisectors, but I am not able to figure out which angle would contain the point $(-1,4)$. Can someone explain how I can figure this out so as to identify the equation of the right bisector the question demands.       
 A: Once you know how to find those bisectors I think the problem is how to locate the point $P=(-1,4)$ in one of those for regions.
Note that you can distinguish those regions looking to up or down of each line:

For example, regions $2$ is above of line $s$ and below of line $r$. Taking as reference the axis $OY$.
First, consider the equation $(s)$ $3x+4y+12=0$. If you take $x=-1$ you get $y=-9/4$ which is below of $y_p=4$. So $P$ is above of the line $3x+4y+12=0$.
Samething for $(r)$ $12x-5y+7=0$. If you take $x=-1$ you get $y=1$ which is below of $y_p=4$. So $P$ is above of the line $12x-5y+7=0$.
So the point $P$ would be in the region $1$.
A: Hint:
If $O$ is the intersection point of the two lines and $P$ the other point, compare the slope of the line $OP$  with the slopes of the two lines (pay attention to vertical lines).
A: You have found the equations of the bisectors $b_1:a_1x+b_1y+c_1=0$ and $b_2:a_2x+b_2y+c_2=0$.
To determine which one of the bisectors contain a point $P=(x_0,y_0)$ you simply insert the point and check that satisfies the equations.
