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If $c_1, c_2$ are positive then the angle bisector of the acute angle between the lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is

  • $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\lt0$

  • $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\gt0$

I wonder what the proof for this is.

I understand the perpendicular distances from a point to the lines are being equated. My doubt is regarding the signs taken.

If the angle between the lines is $\theta$ then the angle between the bisector and either of the lines is $\frac\theta2$.

Using the formula for angle between two lines, if $|\tan\frac\theta2|$ comes out to be less than $1$ then the acute angle is being bisected by the line in concern. Or if $|\tan\frac\theta2|\gt1$ then the obtuse angle.

But how to find that direct relation with $a_1a_2, b_1b_2$?

My Attempt:

Let the slopes of the given lines be $m_1, m_2$. Thus, $m_1=-\frac {a_1}{b_1}$ and $m_2=-\frac {a_2}{b_2}$

If $m_1m_2=-1$ then the lines are perpendicular.

If $m_1m_2\lt-1$ then can we say the angle between them is acute?

If yes, we get $-\frac{a_1}{b_1}\cdot-\frac{a_2}{b_2}\lt-1$, thus, $a_1a_2+b_1b_2\lt0$

But from this, how do we conclude that the angle bisector would be obtained by taking a plus sign in the formula?

aarbee
  • 8,246

1 Answers1

1

If $c_1, c_2$ are positive then the angle bisector of the acute angle between the lines $L_1:a_1x+b_1y+c_1=0$ and $L_2:a_2x+b_2y+c_2=0$ is

$B_1: \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\lt0$

$B_2: \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$, if $a_1a_2+b_1b_2\gt0$

Proof

The $\tan$ of angle between first line $L_1$ and bisector $B_1$ can be worked out as $$|\tan \theta| =\frac{|p|}{\sqrt{p^2+q^2-q}}, \quad p=a_1b_2-b_1a_2, \quad q=a_1a_2+b_1b_2.$$ When $q<0 \implies |\tan \theta|<1 \implies \theta < \pi/4.$ So $B_1$ given by you is acute angle bisector.

Next, if $q>0$, $$\sqrt{p^2+q^2}\le |p|+|q| \implies \sqrt{p^2+q^2}-q\le |p|+|q|-q=|p| \implies |\tan \theta|>1.$$ Then $B_1$ will be obtuse angle bisector.

Z Ahmed
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