Equation of Bisector of Two Lines

[BvU]

I wouldn't worry too much about exams. Doing exercises develops your insight and experience and with that exams are a breeze.

Without a calculator, you can note that $\tan = 3/4$ and $\tan = -12/5$ (especially in combination) remind you of the simplest Pythagorean rectangular triangles 3,4,5 and 12,5,13 !

If you want it spelled out: Let $\tan (\alpha_1/2) = m$ when $\tan (\alpha_1) = 3/4$, then $${2m\over 1-m^2}=3/4 \ \Rightarrow \ m^2 + {8\over 3} m -1 = 0 \\ \Rightarrow \ (m+{4\over 3})^2 - (1+{16\over 9}) = 0 \ \Rightarrow \ m = -{4\over 3} \pm {5\over 3}$$and you want the positive one: $m = {1\over 3}$. Cute, eh ?

The other one yields an equally nice $m_2$ and then $m+ m_2\over 1-m m_2$ is pretty rational too. If you have one, you have the other from $-1/$ and a line with slope p through (-3/7, 10/7) is easily parametrized !

 

[Raghav Gupta]

I wouldn't worry too much about exams. Doing exercises develops your insight and experience and with that exams are a breeze.

Without a calculator, you can note that $\tan = 3/4$ and $\tan = -12/5$ (especially in combination) remind you of the simplest Pythagorean rectangular triangles 3,4,5 and 12,5,13 !

If you want it spelled out: Let $\tan (\alpha_1/2) = m$ when $\tan (\alpha_1) = 3/4$, then $${2m\over 1-m^2}=3/4 \ \Rightarrow \ m^2 + {8\over 3} m -1 = 0 \\ \Rightarrow \ (m+{4\over 3})^2 - (1+{16\over 9}) = 0 \ \Rightarrow \ m = -{4\over 3} \pm {5\over 3}$$and you want the positive one: $m = {1\over 3}$. Cute, eh ?

The other one yields an equally nice $m_2$ and then $m+ m_2\over 1-m m_2$ is pretty rational too. If you have one, you have the other from $-1/$ and a line with slope p through (-3/7, 10/7) is easily parametrized !

Hmm, this is okay
I was searching on google and found a very different approach and fast way
See link ditutor[\URL]
How that formula is derived?

 

[BvU]

Ha, you found something on the internet. At least that means you don't worry about exams that much any more (or do you have internet available then ?)

Now you can memorize that doubtlessly useful expression, or you can develop some insight. We've become familiar enough that I can safely assume the latter of the two. And in an edit you even ask for the derivation. Bear in mind that this is well beyond the call of duty wrt PF: they help with exercises but aren't a substitute for universities, schools or other means of teaching !

Reason I do it anyway is because it so nicely builds on your recently acquired knowledge of the normal equation for a line. (to me that also means that your curriculum and the exercises are well thought-through) !

The derivation: they write down the distances of P to the lines and set them equal. That's all.

To understand, you need an expression for the distance of a point to a line.

Unit normal equation for $Ax+By+C=0$ is $\displaystyle {Ax+By+C\over \sqrt{A^2+B^2}} =0$ .

Distance of line to origin is $|C|\over \sqrt{A^2+B^2}$ .


Unit normal equation for parallel line through $\vec P = (x_P, y_P)$ is $\displaystyle{Ax_P+By_P+D\over \sqrt{A^2+B^2}} =0$.

Distance of line to origin is $\displaystyle|D|\over \sqrt{A^2+B^2}$ .


Distance of P to line is $\displaystyle |D-C| = {|{Ax_P+By_P+C}|\over \sqrt{A^2+B^2} }$. Hey !​

 

[Raghav Gupta]

The derivation: they write down the distances of P to the lines and set them equal. That's all.

Well I am practicing for exam, so net is available to me now but was enchanted by that formula.
I know the distance formula from a line to point.
Understand now, that since it's a angle bisector the distances must be equal.

Thanks and a very much sorry from me.
It must have been painful for you to write that much text.

 

[BvU]

Well I am practicing for exam, so net is available to me now but was enchanted by that formula.
I know the distance formula from a line to point.
Understand now, that since it's a angle bisector the distances must be equal.

Thanks and a very much sorry from me.
It must have been painful for you to write that much text.

No need to apologize: you aren't forcing me, I do it all because I learn a bit too, and it's enjoyable. Good luck with your studies !