Angle Bisectors of Two Lines

Locus of points equidistant from both lines. The + sign gives one bisector, the − sign gives the other.

A point $P(x,y)$ lies on a bisector of lines $L_1: a_1x+b_1y+c_1=0$ and $L_2: a_2x+b_2y+c_2=0$ if and only if it is equidistant from both:

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

The $+$ sign gives one bisector, the $-$ sign gives the other. The two bisectors are always perpendicular to each other.

Which bisector contains the origin?

Substitute $(0,0)$: evaluate $\dfrac{c_1}{\sqrt{a_1^2+b_1^2}}$ and $\dfrac{c_2}{\sqrt{a_2^2+b_2^2}}$.

• Same sign $\Rightarrow$ origin satisfies the $+$ equation $\Rightarrow$ origin lies on the $+$ bisector.
• Opposite sign $\Rightarrow$ origin lies on the $-$ bisector.

Acute vs obtuse bisector

Ensure $c_1, c_2 > 0$ (multiply lines by $-1$ if needed). Then:

• $a_1a_2 + b_1b_2 > 0$: the $+$ bisector bisects the obtuse angle; the $-$ bisector bisects the acute angle.
• $a_1a_2 + b_1b_2 < 0$: the $+$ bisector bisects the acute angle; the $-$ bisector bisects the obtuse angle.