Distinguishing between angular bisectors

  • #1
JC2000
187
15

Homework Statement

:[/B]
The following expression stands for the two angular bisectors for two lines :


[tex]\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}}}\qquad[/tex]

Homework Equations


The equations for the two lines are :
##a_1x + b_1y + c_1 = 0~##and ##a_2x + b_2y + c_2 = 0~##

The Attempt at a Solution


One way is to find the slope from the equation for the respective bisectors and then use the slope of one of the given lines and then apply the formula for angle between two lines to check if the bisector is acute or not. I was wondering if there was a more direct way to do this.

My questions :
(A)
Since the two lines can be bisected in two ways, is there a formula to distinguish between the bisector that bisects the external angle and the bisector that bisects the internal or acute angle between the two lines?
My book says that if [tex] a_1 * a_2 + b_1 * b_2 > 0[/tex] then this refers to the external bisector.
I am stumped by this because : (B) How is this expression arrived at? (C) Since [tex]a_1, a_2, b_1, b_2[/tex] all refer to the original equations of the two lines and not the bisectors, which bisector do this expression refer to?
 
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Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
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Homework Statement

:[/B]
The following expression stands for the two angular bisectors for two lines :


[tex]\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}}}\qquad[/tex]

Homework Equations


The equations for the two lines are :
[tex]ax_1 + by_1 + c_1 = 0[/tex]
and [tex]ax_2 + by_2 + c_2 = 0[/tex]

The Attempt at a Solution


One way is to find the slope from the equation for the respective bisectors and then use the slope of one of the given lines and then apply the formula for angle between two lines to check if the bisector is acute or not. I was wondering if there was a more direct way to do this.

My questions :
(A)
Since the two lines can be bisected in two ways, is there a formula to distinguish between the bisector that bisects the external angle and the bisector that bisects the internal or acute angle between the two lines?
My book says that if [tex] a_1 * a_2 + b_1 * b_2 > 0[/tex] then this refers to the external bisector.
I am stumped by this because : (B) How is this expression arrived at? (C) Since [tex]a_1, a_2, b_1, b_2[/tex] all refer to the original equations of the two lines and not the bisectors, which bisector do this expression refer to?


Did you mean to write the equations of the two lines as
##a_1x + b_1y + c_1 = 0~##and ##a_2x + b_2y + c_2 = 0~##?
 
  • #3
JC2000
187
15
Did you mean to write the equations of the two lines as
##a_1x + b_1y + c_1 = 0~##and ##a_2x + b_2y + c_2 = 0~##?

Yes. I will edit the formatting now...Thanks!
 
  • #4
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
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It turns out that you can find the answer to this problem on several websites with several different approaches. I recommend that you do the research.
 

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