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

The general form of the equation of the angle bisector of two lines is:

$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}= \pm\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$

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).

Edit: $L_1 \equiv a_1x+b_1y+ c_1$

$L_2\equiv a_2x+b_2y+ c_2$

• What are $L_1$ and $L_2$? – Emilio Novati Sep 22 '17 at 9:35
• @EmilioNovati Expressions of Lines – Abcd Sep 22 '17 at 9:42
• Closely related to math.stackexchange.com/q/2403530/265466. – amd Sep 23 '17 at 0:42
• "I have been instructed to do as follows: " Rather than telling us what someone else told you to do, you should show more of what you did do. – Simply Beautiful Art Sep 24 '17 at 18:25
• Thanks! $\ddot\smile$ – Simply Beautiful Art Sep 24 '17 at 18:52

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 $$\frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}} = - \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}.$$

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 $L_1:a_1x+b$): $$\tan\beta=\frac{a_2}{\sqrt{1+a_2^2}-1}$$

• How did you get "$\sin \alpha= 1/\sqrt{1+a^2}$" – Abcd Sep 22 '17 at 10:45
• if You'll take a function y(x)=-(ax+b) (see, that $y(x)+ax+b=0$), you can take the right triangle $(0,y(0)),(1,y(1)),(1,y(0))$ and easily compute $\sin\alpha$. – Jaroslaw Matlak Sep 22 '17 at 11:10