# 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".

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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.)

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.