Distinguishing between angular bisectors

AI Thread Summary
The discussion focuses on distinguishing between the internal and external angular bisectors of two lines represented by their equations. It highlights the formula a_1 * a_2 + b_1 * b_2 > 0, which indicates the external bisector, raising questions about its derivation and application. Participants express confusion regarding how this expression relates to the original line equations rather than the bisectors themselves. The conversation suggests that various resources online provide different methods for solving this problem. Overall, the need for clarity on the relationship between line equations and their bisectors is emphasized.
JC2000
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Homework Statement

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

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 a_1 * a_2 + b_1 * b_2 > 0 then this refers to the external bisector.
I am stumped by this because : (B) How is this expression arrived at? (C) Since a_1, a_2, b_1, b_2 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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JC2000 said:

Homework Statement

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

Homework Equations


The equations for the two lines are :
ax_1 + by_1 + c_1 = 0
and ax_2 + by_2 + 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 a_1 * a_2 + b_1 * b_2 > 0 then this refers to the external bisector.
I am stumped by this because : (B) How is this expression arrived at? (C) Since a_1, a_2, b_1, b_2 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~##?
 
kuruman said:
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!
 
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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