6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This method is not good, what does it say when $c = r = 0$? $\endgroup$ – Gribouillis Aug 23 '17 at 14:11
  • $\begingroup$ Why less than $45$ ? Should not it be less than $90$ ? $\endgroup$ – A---B Aug 23 '17 at 14:12
  • $\begingroup$ @A---B The angle between the two lines should be less than 90. The bisectors divides the angle into two halves. So, the angle between the bisector and a line should be less than 45. $\endgroup$ – Shubhraneel Pal Aug 23 '17 at 14:16
  • $\begingroup$ @Gribouillis for c=r=0, we can still use the method. Suppose we change the sign of $a$ and $b$. Then the sign of ap + bq also changes and so does which sign we use for obtuse and acute. But we are applying this sign change to the formula, where we use the same changed signs for $a$ and $b$. So the net effect is the same equation for acute and obtuse bisectors. $\endgroup$ – Shubhraneel Pal Aug 23 '17 at 14:25
  • $\begingroup$ @ShubhraneelPal My mistake, I read between the lines not between the line and bisector. $\endgroup$ – A---B Aug 23 '17 at 14:31
9
$\begingroup$

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.

enter image description here

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 $PF_1QF_2$.

enter image description here

$\angle{PF_1Q}=\angle{PF_2Q}=\pi/2$, 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.

$\endgroup$
  • $\begingroup$ Nice proof. I think it enlightens the analytic proof. $\endgroup$ – Gribouillis Aug 25 '17 at 8:16
  • 1
    $\begingroup$ @Gribouillis I also tried a related approach that first develops a rule for choosing which of $\|w\|v\pm\|v\|w$ is the acute bisector of a pair of linearly independent vectors $v$, $w$, then applies it to the normals of the lines, and finally works out the acute bisector’s equation from this normal and the intersection point. Working directly with the distances to the lines instead avoids some simple but tedious algebraic manipulations to get to the desired formula. $\endgroup$ – amd Aug 25 '17 at 8:24
  • $\begingroup$ What does signed distance mean? $\endgroup$ – Archer Sep 28 '17 at 14:51
  • $\begingroup$ @Abcd Explained in the rest of that paragraph: the sign of the expression tells you in which half-plane the point lies. $\endgroup$ – amd Sep 29 '17 at 17:10
  • $\begingroup$ how do u know the direction of the perpendicular vectors ?. what if i take the vectors in opposite directions ? $\endgroup$ – ss1729 Aug 1 at 10:20
2
$\begingroup$

Let given $\angle ABC$ and $ax+by+c=0$ one of two equations of bisector $BD$, which you got.

Let $M(x_M,y_M)$ and $f(M)=ax_{M}+by_{M}+c$.

Thus, if $f(A)f(C)<0$ then $BD$ is a bisector of our angle.

If $f(A)f(C)>0$ then it's not so.

The proof of this statement based on continuously of $f$.

For example, if $f(A)>0$ and $f(C)<0$ thus, there is $D\in AC$ for which $f(D)=0$,

which you want, because in this case the ray $BD$ placed between rays $BA$ and $BC$.

There is another way.

If we want to find an equation of the line $BD$, where $BD$ is a bisector of $\Delta ABC$,

then we can use $$\frac{AB}{BC}=\frac{AD}{DC}$$ and to find coordinates of $D$, coordinates of $A$ and to write an equation of $BD$.

It's easier sometimes.

$\endgroup$
  • $\begingroup$ What does "based on continuously of $f$ mean? $\endgroup$ – Archer Sep 28 '17 at 14:55
  • $\begingroup$ @Abcd See please my words in the following line. $\endgroup$ – Michael Rozenberg Sep 28 '17 at 15:16
1
$\begingroup$

I have an analytic proof. It's not particularly pretty, but it seems to work. Let ${L}_{1} , {L}_{2}$ be the two lines and $\binom{{x}_{0}}{{y}_{0}}$ their intersection. By using ${x'} = x-{x}_{0}$ and ${y'} = y-{y}_{0}$ instead of $x$ and $y$, we can suppose that $\boxed{c = r = 0}$.

Then we change nothing to the equation nor to the result by supposing that $\boxed{{a}^{2}+{b}^{2} = {p}^{2}+{q}^{2} = 1}$. The equations of ${L}_{1}$ and ${L}_{2}$ are now $a x+b y = 0$ and $p x+q y = 0$.

The point $M = \binom{a+p}{b+q}$ is on a bissector of ${L}_{1} , {L}_{2}$. In fact its coordinates satisfy $a x+b y = 1+a p+b q = p x+q y$ which is an instance of the $+$ case in your formula. We want to compute the angle ${\alpha}$ between ${L}_{1}$ and the line $O M$.

Let ${M}_{1}$ be the orthogonal projection of $M$ on ${L}_{1}$. Its coordinates are $t \binom{{-b}}{a}$ for some $t \in \mathbb{R}$. We find $t$ by the condition that the vector $\binom{a+p}{b+q}-t \binom{{-b}}{a}$ is proportional to $\binom{a}{b}$. The result is $t = a q-b p$.

We now compute ${M}_{1} \cdot M =-t b \left(a+p\right)+t a \left(b+q\right) = t \left(a q-b p\right) = {t}^{2}$. In the same way, ${\left|M\right|}^{2} = {\left(a+p\right)}^{2}+{\left(b+q\right)}^{2} = 2 \left(1+a p+b q\right)$. Finally $\left|{M}_{1}\right| = \left|t\right|$. It follows that

$$\left|\cos \left({\alpha}\right)\right| = \frac{{M}_{1} \cdot M}{\left|M\right| \left|{M}_{1}\right|} = \frac{\sqrt{2}}{2} \frac{\left|t\right|}{\sqrt{1+a p+b q}}$$

If $a p+b q \geqslant 0$, then $\left|t\right| \leqslant 1 \leqslant \sqrt{1+a p+b q}$ and it follows that $\left|\cos \left({\alpha}\right)\right| \leqslant \frac{\sqrt{2}}{2}$. Hence we have ${\alpha} \in \left[\frac{{\pi}}{4} , \frac{{\pi}}{2}\right]$, which means that $2 {\alpha} \in \left[\frac{{\pi}}{2} , {\pi}\right]$ and the point $M$ is on the obtuse bissector.

This proves your book's claim. To study the negative sign, you can use the point ${M'} = \binom{a-p}{b-q}$ on the second bissector line.

Note that the signs of $c$ and $r$ are not used at all in the proof. You can safely remove them from the condition.

Edit: it appears to me now that the proof can be further shortened by not computing the orthogonal projection $M_1$. One can use $\binom{-b}{a}$ directly to compute the angle.

Edit 2: A better proof

Lemma 1 Let $x \in \left({-1} , 1\right]$ and $y \in \left[{-1} , 1\right]$ be such that ${x}^{2}+{y}^{2} = 1$ and let

$${\theta} = \arctan \left(\frac{y}{1+x}\right)$$

then one has $\cos \left(2 {\theta}\right) = x$ and $\sin \left(2 {\theta}\right) = y$.

proof Use the classical formulas, with $t = \tan \left({\theta}\right)$

$$\cos \left(2 {\theta}\right) = \frac{1-{t}^{2}}{1+{t}^{2}} \quad \text{ and } \quad \sin \left(2 {\theta}\right) = \frac{2 t}{1+{t}^{2}}$$

Let now our normal unit vectors be ${n}_{1} = \binom{a}{b}$ and ${n}_{2} = \binom{p}{q}$. We assume that ${n}_{1} \cdot {n}_{2} = a p+b q \geqslant 0$. We define

$${\theta} = \arctan \left(\frac{a q-b p}{1+a p+b q}\right)$$

It follows from the lemma that $\cos \left(2 {\theta}\right) = a p+b q = {n}_{1} \cdot {n}_{2}$ and $\sin \left(2 {\theta}\right) = a q-b p$. One has ${\theta} \in \left[{-\frac{{\pi}}{4}} , \frac{{\pi}}{4}\right]$ because $\cos \left(2 {\theta}\right) \geqslant 0$. We ensure that ${\theta} \neq 0$ by requiring that ${n}_{1}$ and ${n}_{2}$ are not proportionals. Let us define the vectors

$${e}_{1} = \frac{1}{2 \sin \left({\theta}\right)} \left({n}_{1}-{n}_{2}\right) \quad \text{ and } \quad {e}_{2} = \frac{1}{2 \cos \left({\theta}\right)} \left({n}_{1}+{n}_{2}\right)$$

The calculus shows easily that $\left({e}_{1} , {e}_{2}\right)$ are a direct orthonormal basis and that

$${n}_{1} =\sin \left({\theta}\right) {e}_{1}+\cos \left({\theta}\right) {e}_{2} \quad \text{ and } \quad {n}_{2} = -\sin \left({\theta}\right) {e}_{1}+\cos \left({\theta}\right) {e}_{2}$$

It follows that the equations of ${L}_{1}$ and ${L}_{2}$ in the basis $\left({e}_{1} , {e}_{2}\right)$ are respectively $y = -\tan \left({\theta}\right) x$ and $y =\tan \left({\theta}\right) x$ which means that ${L}_{1} , {L}_{2}$ are the straight lines with slopes $-\tan \left({\theta}\right)$ and ${\tan } \left({\theta}\right)$. The angle between the lines in the direction of ${e}_{1}$ is therefore $2 {\theta} \in \left[{-\frac{{\pi}}{2}} , \frac{{\pi}}{2}\right]$, i.e. an acute angle, as the book claims.

A further consequence of this proof is that if $a q-b p > 0$, then the rotation from ${L}_{1}$ to ${L}_{2}$ in the acute sector goes in the anticlockwise direction, and if $a q-b p < 0$, this rotation goes in the clockwise direction.

$\endgroup$
  • $\begingroup$ You might be able to simplify this a bit by using the fact that $(a\pm p,b\pm q)^T$ are the diagonals of a parallelogram with sides defined by the two normals. $\endgroup$ – amd Aug 25 '17 at 17:16
  • $\begingroup$ @amd Indeed, it is a rhombus which sides are orthogonal to the two lines, but I have found a new proof which I find better. $\endgroup$ – Gribouillis Aug 26 '17 at 10:01
  • 1
    $\begingroup$ Your “better proof” also needs to consider the case $\theta=\pi$ for completeness. $\endgroup$ – amd May 7 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.