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".

  • $\begingroup$ The values of $c_1$, $c_2$ play no rôle in this problem. It is sufficient to discuss the case $c_1=c_2=0$. If the rule you have found is o.k. in this case it is o.k. always. $\endgroup$ – Christian Blatter Apr 12 '16 at 11:54

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 $$\cos\theta=\frac{\left|a_1\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)+b_1\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)\right|}{\sqrt{a_1^2+b_1^2}\sqrt{\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)^2+\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)^2}}$$


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.)

  • $\begingroup$ Great answer Sir !!! I wonder why OP hasn't up voted it the best answer. $\endgroup$ – Mathematics Aug 28 '16 at 11:35
  • $\begingroup$ How did you find the cosine of angle between the two lines? $\endgroup$ – Archer Sep 29 '17 at 3:02
  • $\begingroup$ @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 $$\cos\theta=\frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}$$ $\endgroup$ – mathlove Sep 29 '17 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.