The formula for finding the angular bisectors of two lines $ax+by+c=0$ and $px+qy+r=0$ is $$\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm\frac{px+qy+r}{\sqrt{p^2+q^2}}$$
I understand the proof of this formula but I do not understand how to determine which sign is for acute bisector and which one for obtuse.
I can find the angle between a bisector and a line, and if it comes less than $45^\circ$ then it is acute bisector.
But that is a lengthy method and involves calculation.
My book says, if $ap+bq$ is positive then the negative sign in the formula is for acute bisector.
I want a proof of this method.
Edit: Using the method for finding the position of two points with respect to a line is okay for the proof.
[Math] How to prove this method of determining the sign for acute or obtuse angle bisector in the angle bisector formula works
analytic geometrygeometry
Best Answer
Recall that for the line given by the equation $\lambda x+\mu y+\tau=0$, the vector $(\lambda,\mu)$ is normal to the line and that $d={\lambda x+\mu y+\tau\over\sqrt{\lambda^2+\mu^2}}$ is the signed distance of the point with coordinates $(x,y)$ from the line: if it is positive, the normal vector points from the line toward the point, while a negative value means that the normal points away from the point. Another way to put this is that the sign of $d$ tells us in which half-plane the point lies.
A pair of distinct intersecting lines $ax+by+c=0$ and $px+qy+r=0$ divides the plane into four regions, which we can number I-IV counterclockwise, beginning with the one into which both of the normals $\mathbf n_1=(a,b)$ and $\mathbf n_2=(p,q)$ point, as illustrated below.
The distances from the two lines have the same sign for points in regions I and III, and opposite signs in regions II and IV. Now, let $P$ be a point in the interior of region I and $Q$ the intersection of the two lines. Drop perpendiculars from $P$ meeting the lines at points $F_1$ and $F_2$ and consider the quadrilateral $PF_1QF_2$.
$\angle{PF_1Q}=\angle{PF_2Q}=\pi/2$, therefore $\angle{F_1PF_2}$ and $\angle{F_1QF_2}$ are complementary. $\angle{F_1PF_2}$ is equal to the angle between the normal vectors, so if this angle is acute, then $\angle{F_1QF_2}$ is obtuse and the acute angle bisector of the two lines lies in regions II and IV; if the angle between the normals is obtuse, then the bisector runs through regions I and III. We can determine the type of angle formed by the two normals by examining the sign of their dot product $ap+bq$: if this is positive, the angle is acute; if negative, obtuse. (If zero, they are orthogonal, so there’s no acute angle bisector per se.)
Points along an angle bisector of two lines are equidistant from the lines. Using the formula for the distance from a line from the top of this answer, this means that an equation for the acute angle bisector is $${ax+by+c\over\sqrt{a^2+b^2}}=\pm{px+qy+r\over\sqrt{p^2+q^2}},$$ with the sign on the right-hand side chosen opposite to that of $ap+bq$.
Notice that no restrictions were placed on $c$ and $r$ in the above construction, so arranging for these values to be positive is not necessary. Also, this means that the construction works even when either or both of these constant terms is zero.