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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".

Matt
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2 Answers2

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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
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    Great answer Sir !!! I wonder why OP hasn't up voted it the best answer. – Mathematics Aug 28 '16 at 11:35
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    How did you find the cosine of angle between the two lines? – Archer Sep 29 '17 at 03:02
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    @Abcd: If $\theta$ is the (smaller) angle between $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, then we have $$\cos\theta=\frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$ – mathlove Sep 29 '17 at 06:30
  • @mathlove, Excellent answer! Could you please explain, how "$c_1,c_2$ both are of same sign" is irrelevant here? I think the sign of the expression $a_1a_2+b_1b_2$ depends on the fact mentioned and is relevant to our study. – Vishnu Oct 18 '19 at 12:59
  • @Intellex: The sign of $a_1a_2+b_1b_2$ is irrelevant to "$c_1,c_2$ both are of same sign", isn't it? – mathlove Oct 18 '19 at 16:28
  • @mathlove, Please see my answer below and point out any mistakes (if any). – Vishnu Oct 19 '19 at 03:31
  • @mathlove: but what you've used for $cos\theta$ doesn't seem like $\frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$. At the end, it looks like it simplifies to $sin\frac \theta 2$ – harry Apr 23 '21 at 05:28
  • @Harry Holmes : Note that $a_1,a_2,b_1,b_2$ written in my comment above are different from $a_1,a_2,b_1,b_2$ written in my answer. – mathlove Apr 23 '21 at 10:54
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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.

Vishnu
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  • When you change $a_2x+b_2y-q=0$ to $-a_2x-b_2y+q=0$, you actually just change $L_+$ to $L_-$. Please remember $$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$ – mathlove Oct 19 '19 at 15:34
  • @mathlove, I understand that. But the sign of $a_1a_2+b_1b_2$ also changes which affects the fact whether it represents the obtuse or acute angle bisector as in your answer. – Vishnu Oct 19 '19 at 15:37
  • Let us consider the case $r\gt 0$. When you change $a_2x+b_2y-q=0$ to $-a_2x-b_2y+q=0$, we have $r\lt 0$, and then $L_-$ is the line we want. No problem in my answer. – mathlove Oct 19 '19 at 15:43
  • Sorry @mathlove, if you misunderstood me. I did not mean to point out any problem in your answer. I just wanted to know whether my book is right. I am unable to understand your point. It would be great if you could explain this further. I think the sign of the constants matter when they are of opposite signs but not during the case they have the same signs. – Vishnu Oct 20 '19 at 02:38
  • When the constant terms are of opposite signs, the sign of the expression $a_1a_2+b_1b_2$ depends on which form of equation we choose. In Case-II of my answer even though $-a_1a_2-b_1b_2=-r$ is equivalent to $a_1a_2+b_1b_2=r$, in reality they are not. It must be noted $-a_1$ and $-a_2$ are new $a_1$ and $a_2$. I have not introduced any new variables, just kept on going with the old set. – Vishnu Oct 20 '19 at 02:46
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    Let me add another explanation. (1) If you consider $5x+12y+1=0$ and $3x+4y-1=0$, then $L_+$ is the obtuse angle bisector. (2) If you consider $5x+12y+1=0$ and $-3x-4y+1=0$, then $L_-$ is the obtuse angle bisector. (3) The sign of the constants does not matter even when they are of opposite signs. "$c_1$ and $c_2$ both are of the same sign" is irrelevant. – mathlove Oct 20 '19 at 04:36
  • Thank you @mathlove :) Now I understood. However, I am not going to remove my answer to preserve your comments. – Vishnu Oct 20 '19 at 04:41