How to find whether equation of angle bisector represents the obtuse or acute angle bisector of two given straight lines?

by Matt   Last Updated October 20, 2019 05:20 AM - source

Two lines: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are given. I know that the equation of its bisectors is ${a_1x + b_1y + c_1 \over \sqrt{(a_1^2 + b_1^2)}} = \pm {a_2x + b_2y + c_2 \over\sqrt{ (a_2^2 + b_2^2)}}$ But I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming $c_1 , c_2$ both are of same sign, I know if $a_1a_2 + b_1b_2 > 0$ and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. $tan θ = {m_1 - m_2 \over 1+ m_1m_2}$ and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming $c_1 , c_2$ both are of same sign IF $a_1a_2 + b_1b_2 > 0 $then if we take positive sign we get the obtuse angle bisector".



Answers 2


We have two lines : $$L_1 : a_1x+b_1y+c_1=0,\quad L_2 : a_2x+b_2y+c_2=0$$

and the angle bisectors : $$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$

If we let $\theta$ be the (smaller) angle between $L_+$ and $L_1$, then we have $$\cos\theta=\frac{\left|a_1\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)+b_1\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)\right|}{\sqrt{a_1^2+b_1^2}\sqrt{\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)^2+\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)^2}}$$

$$=\frac{\left|\sqrt{a_1^2+b_1^2}-\frac{a_1a_2+b_1b_2}{\sqrt{a_2^2+b_2^2}}\right|}{\sqrt{a_1^2+b_1^2}\sqrt{2-2\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}}\times\frac{2\frac{1}{\sqrt{a_1^2+b_1^2}}}{2\frac{1}{\sqrt{a_1^2+b_1^2}}}=\sqrt{\frac{1-\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}{2}}$$

Hence, we can see that $$\begin{align}a_1a_2+b_1b_2\gt 0&\iff\cos\theta\lt 1/\sqrt 2\\&\iff \theta\gt 45^\circ\\&\iff \text{$L_+$ is the obtuse angle bisector}\end{align}$$ as desired.

(Note that "$c_1,c_2$ both are of same sign" is irrelevant.)

mathlove
mathlove
April 12, 2016 16:40 PM

This answer is preserved for those who want to understand why the sign of constants do not matter. After reading this answer please check the comments for more details from @mathlove.

@mathlove's answer really explains the question. But, I would like to show that the "$c_1,c_2$ are of same sign" is relevant to our study here, unlike @mathlove's answer.

This is my hypothesis:

Consider two lines represented by $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$. Based on the nature of signs of $c_1$ and $c_2$, we have two cases:

Case I: Both $c_1$ and $c_2$ are of same sign:

$a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ can also be represented as (multiplying both sides by $-1$) $-a_1x-b_1y-c_1=0$ and $-a_2x-b_2y-c_2=0$ respectively. In both, the original equation and the negated equation the sign of $a_1a_2+b_1b_2$ remains the same. So, "$c_1,c_2$ are of same sign" seems to be irrelevant.

Now consider,

Case II: Both $c_1$ and $c_2$ are of opposite signs:

Let us consider $c_1=+p$ and $c_2=-q$ where $p$ and $q$ are positive real numbers.

So, $a_1x+b_1y+p=0$ and $a_2x+b_2y-q=0$ are the equations of the lines under consideration. Let $a_1a_2+b_1b_2=r$ where $r$ is any real number, positive or negative.

The equation of the second line can also be represented as $-a_2x-b_2y+q=0$ by multiplying by $-1$ on both sides. Now, $-a_1a_2-b_1b_2=-r$ clearly of opposite sign compared to the previous form.

Conclusion:

"$c_1,c_2$ are of same sign (or of opposite sign)" is relevant to our study here.

Intellex
Intellex
October 19, 2019 03:30 AM

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