Equation of the bisector of the angle between two lines containing the given point

Question

The general form of the equation of the angle bisector of two lines : $$ \begin{align}L_1 &=a_1x+b_1y+ c_1=0 \\L_2 &= a_2x+b_2y+ c_2=0 \end{align}$$ Given as: $$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$

Now, we have a point $(\alpha, \beta)$ lying in one of the angles between these two.

I have been instructed to do as follows:

1. Find the sign of the expression of $L_1$ and $L_2$.
2. If they are opposite then, choose the negative sign from the general form otherwise choose the positive

but I want to know this method's proof for better understanding. Or, if there's an alternate/ better method to deal with it, please let me know (with proof).

Answer 1

I'd use other method (for lines $y+ax+b=0$).

$L: y+ax+b=0$ creates with ox axis the angle $\alpha$, then $$\sin\alpha=\frac{a}{\sqrt{1+a^2}}\\ \cos\alpha=\frac{1}{\sqrt{1+a^2}}$$

If two lines $L_1, L_2$ are not paralel ($a_1\neq a_2$), we can compute their intersection point $(x_0, y_0)$

The angle between $L_1$ and $L_2$ is $\beta=\frac{\alpha_1+\alpha_2}{2}$

Thus:

• if $\alpha_1+\alpha_2=\pi(1+2k)$ (it can be obtained only, if $a_1=-a_2$), then $\beta=\frac{\pi}{2}+k\pi$ and the bisector line is in form $$L_3:y+b_3=0$$
• in other cases ($a_1\neq-a_2$) we can compute $\tan \beta$: $$\tan\beta = \frac{\sin(\alpha_1+\alpha_2)}{1-\cos(\alpha_1+\alpha_2)}=\frac{\sin(\alpha_1)\cos(\alpha_2)+\sin(\alpha_2)\cos(\alpha_1)}{1+\sin(\alpha_1)\sin(\alpha_2)-\cos(\alpha_1)\cos(\alpha_2)}\\ =\frac{a_1+a_2}{\sqrt{(1+a_1^2)(1+a_2^2)}+a_1a_2-1}$$ Thus the bisector is in the form $$L_3:y-\frac{a_1+a_2}{\sqrt{(1+a_1^2)(1+a_2^2)}+a_1a_2-1}x+b_3=0$$

Now all we need is to insert our intersection point into adequate form of $L_3$ to compute $b_3$.

Edit: In case when one of the lines is in form $L=ax+b$:

We have $\sin\alpha=0$, $\cos\alpha=1$ and $\alpha=0$

Lines are paralel, if both are in this form.

If not, computation of $\tan\beta$ gives us different value (for $L_1:a_1x+b$): $$\tan\beta=\frac{a_2}{\sqrt{1+a_2^2}-1}$$

Answer 2

The formula $$ d_1(x,y) = \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} $$ gives you the distance from a point $(x,y)$ to the line $L_1$ whose equation is $a_1x+b_1y+c_1 = 0.$ See Distance Between A Point And A Line for a proof of this. For the line $L_2$ given by $a_2x+b_2y+c_2 = 0,$ the distance of a point to the line is given by $$ d_2(x,y) = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$

A point $(x,y)$ on an angle bisector between two lines is equidistant from the two lines, that is, it satisfies the condition $d_1(x,y) = d_2(x,y).$ Writing out the formulas for $d_1$ and $d_2$ in full, $$ \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$

Now observe that $\lvert a_1x+b_1y+c_1 \rvert$ will be either $a_1x+b_1y+c_1$ or $-(a_1x+b_1y+c_1),$ whichever of those two expressions is positive. In fact, $a_1x+b_1y+c_1$ will be positive for all points on one side of the line and negative for all points on the other side.

Now if the lines $L_1$ and $L_2$ intersect, they divide the plane into four regions. Label each these regions as $+L_1$ or $-L_1$ depending on whether $a_1x+b_1y+c_1$ is (respectively) positive or negative in that region. Label each region as $+L_2$ or $-L_2$ depending on whether $a_2x+b_2y+c_2$ is (respectively) positive or negative in that region.

One of the angle bisectors of $L_1$ and $L_2$ will go through the regions labeled $+L_1,+L_2$ or $-L_1,-L_2.$ That is, on that line the signs of $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$ are either both positive or both negative. Points on this line therefore satisfy the formula $$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}} = \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$ (For points in the region $-L_1,-L_2,$ this formula gives negative values on both sides, but their absolute values are equal.)

The other angle bisector goes through $+L_1,-L_2$ and $-L_1,+L_2$ and has the formula $$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}} = - \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$

Answer 3

There is a much more elegant and geometric view point for deriving this equation. Let the lines be $L_1 = a_1 x + b_1 y + c_1=0$ and $L_2 = a_2 x + b_2 y + c_2=0$, the unit normal for these lines are given as :

$$ \begin{align} n_1 &= \frac{\nabla L_1}{|\nabla L_1|} \\ n_2 &= \frac{|\nabla L_2|}{|\nabla L_2|} \end{align}$$

Now, the geometric interpretation of subtract two unit vectors is finding the vector which is equi-angle from the two unit vectors similar proof i.e: midpoint of the arc of the two vectors.

$$ \chi= n_1 -n_2$$

The above is the vector in direction to the midpoint of the arc cut by the two unit vectors on the unit circle. It is also the normal of one of the bisectors. How do we in the other normal? We rotate it by ninty degrees.

If we denote $\phi(v)$ to be the map which associates a vector with components $(v_x,v_y)$ in the cartesian plane to a complex number with components $z=x+iy$, then we can rotate the previously obtained vector by ninty degrees to find the normal of the other bisector (Relevant theorem):

$$ \tilde{\chi} = \phi^{-1} \left[ i\phi(\chi) \right]$$

Hence, we can write the equation of two bisector as:

$$ \begin{align} \chi \cdot (x-\alpha,y -\beta) &=0 \\ \tilde{\chi} \cdot (x-\alpha,y- \beta) &=0\end{align}$$

Now, check the angle between $n_1$ and $n_2$, if it is greater than ninty degrees then it means $\chi$ is the obtuse bisector and if not, it $\chi$ is the acute bisector. To understand why, check the linked in equi-angle form.