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Regarding an answer on angular bisectors in 2D coordinate geometry.

How do I prove this method of determining the sign for acute or obtuse angle bisector in the angle bisector formula works?Given two lines in 2D, how to select the angle bisector related to the smallest angle between the lines?Equation of angle bisector, given the equations of two lines in 2D$A$ is a point on either of the two lines $y+sqrt3|x|=2$ at a distance of $frac4sqrt3$ units from their point of intersection.Equations of the two lines through the origin which intersect the line $fracx-32=fracy-31=fracz1$ at an angle of $fracpi3$How to find whether equation of angle bisector represents the obtuse or acute angle bisector of two given straight lines?Proof that if two lines are parallel then $A_1$ = $A_2$ and $B_1$ = $B_2$?Bisectors of a triangle meet at point.How do I prove this method of determining the sign for acute or obtuse angle bisector in the angle bisector formula works?Find the equations of the bisectors of the angles between the lines $y=x$ and $y=7x+4$.Find the angle between two tangents from an external point $(x_1,y_1)$ to the circle $x^2+y^2=a^2$ .Need proof for a property in Analytic Geometry

2

$begingroup$

I am aware that the following expression represents the two angular bisectors for two straight lines.

$$fraca_1x+b_1y-c_1sqrta_1^2+b_1^2=pm fraca_2x+b_2y-c_2sqrta_2^2+b_2^2qquad $$

I had the following question which was already asked and answered here:

Is there any link between the sign of RHS and the bisector of the smallest (biggest) angle?

I then had the following questions regarding the best answer:

Forget about $c_1$ and $c_2$, put $u:=(a_1,b_1)/sqrta_1^2+b_1^2$, $v:=(a_2,b_2)/sqrta_2^2+b_2^2$ and let $z:=(x,y)$. The lines $ucdot z=0$ and $vcdot z =0$ are parallel to your lines $g_1$ and $g_2$. If $ucdot v>0$ (i.e., $u$ and $v$ enclose an acute angle) then it easy to see that $u+v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so this bisector is parallel to the line $(u+v)cdot z=0$. If, on the other hand, $ucdot v<0$ then $u$ and $-v$ enclose an acute angle; therefore $u-v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so in this case the desired bisector is parallel to the line $(u-v)cdot z=0$.

1. What are $u$ and $v$ and $z$? Hence what would $u+v$ and $u-v$ be? (I am unable to understand the notation.)


2. How are the dot products parallel to the original lines? (Probably answerable from 1.)


3. Is there a way to use the original equations to tell if a bisector is 'internal' or 'external'?

Apologies for the silly question and for violating any rules. I do not have enough reputation to ask in the comments.

analytic-geometry

share|cite|improve this question

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: 3: the sign of $ucdot v$ is equal to the sign of $(a_1,b_1)cdot(a_2,b_2)$.
  $endgroup$
  – amd
  Mar 12 at 6:05
  

  
  
  
  

add a comment | 

2

$begingroup$

I am aware that the following expression represents the two angular bisectors for two straight lines.

$$fraca_1x+b_1y-c_1sqrta_1^2+b_1^2=pm fraca_2x+b_2y-c_2sqrta_2^2+b_2^2qquad $$

I had the following question which was already asked and answered here:

Is there any link between the sign of RHS and the bisector of the smallest (biggest) angle?

I then had the following questions regarding the best answer:

Forget about $c_1$ and $c_2$, put $u:=(a_1,b_1)/sqrta_1^2+b_1^2$, $v:=(a_2,b_2)/sqrta_2^2+b_2^2$ and let $z:=(x,y)$. The lines $ucdot z=0$ and $vcdot z =0$ are parallel to your lines $g_1$ and $g_2$. If $ucdot v>0$ (i.e., $u$ and $v$ enclose an acute angle) then it easy to see that $u+v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so this bisector is parallel to the line $(u+v)cdot z=0$. If, on the other hand, $ucdot v<0$ then $u$ and $-v$ enclose an acute angle; therefore $u-v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so in this case the desired bisector is parallel to the line $(u-v)cdot z=0$.

1. What are $u$ and $v$ and $z$? Hence what would $u+v$ and $u-v$ be? (I am unable to understand the notation.)


2. How are the dot products parallel to the original lines? (Probably answerable from 1.)


3. Is there a way to use the original equations to tell if a bisector is 'internal' or 'external'?

Apologies for the silly question and for violating any rules. I do not have enough reputation to ask in the comments.

analytic-geometry

share|cite|improve this question

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: 3: the sign of $ucdot v$ is equal to the sign of $(a_1,b_1)cdot(a_2,b_2)$.
  $endgroup$
  – amd
  Mar 12 at 6:05
  

  
  
  
  

add a comment | 

$begingroup$

I am aware that the following expression represents the two angular bisectors for two straight lines.

$$fraca_1x+b_1y-c_1sqrta_1^2+b_1^2=pm fraca_2x+b_2y-c_2sqrta_2^2+b_2^2qquad $$

I had the following question which was already asked and answered here:

Is there any link between the sign of RHS and the bisector of the smallest (biggest) angle?

I then had the following questions regarding the best answer:

Forget about $c_1$ and $c_2$, put $u:=(a_1,b_1)/sqrta_1^2+b_1^2$, $v:=(a_2,b_2)/sqrta_2^2+b_2^2$ and let $z:=(x,y)$. The lines $ucdot z=0$ and $vcdot z =0$ are parallel to your lines $g_1$ and $g_2$. If $ucdot v>0$ (i.e., $u$ and $v$ enclose an acute angle) then it easy to see that $u+v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so this bisector is parallel to the line $(u+v)cdot z=0$. If, on the other hand, $ucdot v<0$ then $u$ and $-v$ enclose an acute angle; therefore $u-v$ is orthogonal to the bisector of the smaller angle between $g_1$ and $g_2$, so in this case the desired bisector is parallel to the line $(u-v)cdot z=0$.

1. What are $u$ and $v$ and $z$? Hence what would $u+v$ and $u-v$ be? (I am unable to understand the notation.)


2. How are the dot products parallel to the original lines? (Probably answerable from 1.)


3. Is there a way to use the original equations to tell if a bisector is 'internal' or 'external'?

Apologies for the silly question and for violating any rules. I do not have enough reputation to ask in the comments.

analytic-geometry

share|cite|improve this question

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

$endgroup$

share|cite|improve this question

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

share|cite|improve this question

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

edited Mar 11 at 12:01

MarianD

9931612

edited Mar 11 at 12:01

MarianD

9931612

asked Mar 11 at 11:45

JC2000JC2000

418

asked Mar 11 at 11:45

JC2000JC2000

418

1 Answer
1

active

oldest

votes

1

$begingroup$

Ad $1$:

$u, v$ are unit vectors perpendicular to those lines. $z := (x, y)$ is a $formal$ vector, so for example

$$u cdot z = 0 $$

is an equation of a line, because it is

$$(u_1, u_2) cdot(x,y) = 0$$
or
$$u_1x+u_2y=0$$
$(u+v)$ and $(u-v)$ are perpendicular to bisectors $f, g$ of original lines $m,n$:

Ad $2$:

As you guessed, it follows from the previous part.

Ad $3$:

Yes, there is. Follow the link in your own question to learn how.

share|cite|improve this answer

edited 2 days ago

answered Mar 11 at 12:35

MarianDMarianD

9931612

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The diagram does help tremendously as I am trying to work my head around 1 and 2. But how does Ad.3 tell us which equation refers to the 'internal' bisector and which to the 'external'? So when '$+$' is used you get one of the bisectors and the other is obtained when '$-$' is used, but which one bisects the acute and which one the obtuse?
  $endgroup$
  – JC2000
  Mar 11 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't tell us, I showed that it is impossible. The problem is that the same line has (infinitely) many different equations.
  $endgroup$
  – MarianD
  Mar 11 at 13:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding 1 and 2. : Since $u,v$ are perpendicular to $m,n$ thus the vector $u+v$ (which is the acute bisector for $u,v$) is the 'obtuse' bisector for the original pair of lines. Thus $(u+v)cdot z=0$ which is perpendicular to $u+v$ is the 'acute bisector' for the original lines?! Sorry for being painfully slow but why is $ucdot z = 0$ parallel to the original?
  $endgroup$
  – JC2000
  Mar 11 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See, you put a question and I answered it. You didn't accept my answer. It seems that you are not satisfied with it, so you have wait for a better one.
  $endgroup$
  – MarianD
  Mar 11 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for being a bit dense but can I ask follow up questions?
  $endgroup$
  – JC2000
  Mar 11 at 16:44
  

  
  
  
  

 | 
show 3 more comments

Your Answer

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1

$begingroup$

Ad $1$:

$u, v$ are unit vectors perpendicular to those lines. $z := (x, y)$ is a $formal$ vector, so for example

$$u cdot z = 0 $$

is an equation of a line, because it is

$$(u_1, u_2) cdot(x,y) = 0$$
or
$$u_1x+u_2y=0$$
$(u+v)$ and $(u-v)$ are perpendicular to bisectors $f, g$ of original lines $m,n$:

Ad $2$:

As you guessed, it follows from the previous part.

Ad $3$:

Yes, there is. Follow the link in your own question to learn how.

share|cite|improve this answer

edited 2 days ago

answered Mar 11 at 12:35

MarianDMarianD

9931612

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The diagram does help tremendously as I am trying to work my head around 1 and 2. But how does Ad.3 tell us which equation refers to the 'internal' bisector and which to the 'external'? So when '$+$' is used you get one of the bisectors and the other is obtained when '$-$' is used, but which one bisects the acute and which one the obtuse?
  $endgroup$
  – JC2000
  Mar 11 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't tell us, I showed that it is impossible. The problem is that the same line has (infinitely) many different equations.
  $endgroup$
  – MarianD
  Mar 11 at 13:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding 1 and 2. : Since $u,v$ are perpendicular to $m,n$ thus the vector $u+v$ (which is the acute bisector for $u,v$) is the 'obtuse' bisector for the original pair of lines. Thus $(u+v)cdot z=0$ which is perpendicular to $u+v$ is the 'acute bisector' for the original lines?! Sorry for being painfully slow but why is $ucdot z = 0$ parallel to the original?
  $endgroup$
  – JC2000
  Mar 11 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See, you put a question and I answered it. You didn't accept my answer. It seems that you are not satisfied with it, so you have wait for a better one.
  $endgroup$
  – MarianD
  Mar 11 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for being a bit dense but can I ask follow up questions?
  $endgroup$
  – JC2000
  Mar 11 at 16:44
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

Ad $1$:

$u, v$ are unit vectors perpendicular to those lines. $z := (x, y)$ is a $formal$ vector, so for example

$$u cdot z = 0 $$

is an equation of a line, because it is

$$(u_1, u_2) cdot(x,y) = 0$$
or
$$u_1x+u_2y=0$$
$(u+v)$ and $(u-v)$ are perpendicular to bisectors $f, g$ of original lines $m,n$:

Ad $2$:

As you guessed, it follows from the previous part.

Ad $3$:

Yes, there is. Follow the link in your own question to learn how.

share|cite|improve this answer

edited 2 days ago

answered Mar 11 at 12:35

MarianDMarianD

9931612

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The diagram does help tremendously as I am trying to work my head around 1 and 2. But how does Ad.3 tell us which equation refers to the 'internal' bisector and which to the 'external'? So when '$+$' is used you get one of the bisectors and the other is obtained when '$-$' is used, but which one bisects the acute and which one the obtuse?
  $endgroup$
  – JC2000
  Mar 11 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't tell us, I showed that it is impossible. The problem is that the same line has (infinitely) many different equations.
  $endgroup$
  – MarianD
  Mar 11 at 13:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding 1 and 2. : Since $u,v$ are perpendicular to $m,n$ thus the vector $u+v$ (which is the acute bisector for $u,v$) is the 'obtuse' bisector for the original pair of lines. Thus $(u+v)cdot z=0$ which is perpendicular to $u+v$ is the 'acute bisector' for the original lines?! Sorry for being painfully slow but why is $ucdot z = 0$ parallel to the original?
  $endgroup$
  – JC2000
  Mar 11 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See, you put a question and I answered it. You didn't accept my answer. It seems that you are not satisfied with it, so you have wait for a better one.
  $endgroup$
  – MarianD
  Mar 11 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for being a bit dense but can I ask follow up questions?
  $endgroup$
  – JC2000
  Mar 11 at 16:44
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Ad $1$:

$u, v$ are unit vectors perpendicular to those lines. $z := (x, y)$ is a $formal$ vector, so for example

$$u cdot z = 0 $$

is an equation of a line, because it is

$$(u_1, u_2) cdot(x,y) = 0$$
or
$$u_1x+u_2y=0$$
$(u+v)$ and $(u-v)$ are perpendicular to bisectors $f, g$ of original lines $m,n$:

Ad $2$:

As you guessed, it follows from the previous part.

Ad $3$:

Yes, there is. Follow the link in your own question to learn how.

share|cite|improve this answer

edited 2 days ago

answered Mar 11 at 12:35

MarianDMarianD

9931612

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Mar 11 at 12:35

MarianDMarianD

9931612

answered Mar 11 at 12:35

MarianDMarianD

9931612

answered Mar 11 at 12:35

MarianDMarianD

9931612