Angle-bisector equations and equidistant loci

Content updated: October 22, 2026

The two angle-bisector lines of two intersecting lines form the locus of points equidistant from the lines. Their equation has two signs, but the plus sign does not always identify the internal bisector of a chosen angle.

On this page: The idea · Method · Examples · Practice · Further learning

1. The locus of points equidistant from two lines

Let $d_i:a_ix+b_iy+c_i=0$, with $a_i^2+b_i^2>0$ for $i=1,2$, and suppose that the lines intersect. A point $M(x;y)$ lies on one of their angle-bisector lines exactly when $d(M,d_1)=d(M,d_2)$. By the distance formula, this is equivalent to

Denote the signed expressions on the two sides by $f_1,f_2$. The equality $|f_1|=|f_2|$ is equivalent to $f_1=f_2$ or $f_1=-f_2$. Hence exactly two angle-bisector lines result.

2. The two signs and selecting a bisector

Both lines pass through the intersection of $d_1,d_2$, and they are perpendicular. Indeed, let $\vec p,\vec q$ be unit normal vectors of the original lines. Normal vectors of the bisectors are $\vec p-\vec q$ and $\vec p+\vec q$; their dot product is $\|\vec p\|^2-\|\vec q\|^2=0$. Since the original lines intersect, both vectors differ from $\vec0$.

To call a bisector internal or external, specify the angle or the two rays under consideration. Multiplying either original equation by $-1$ exchanges the roles of the two signs without changing the pair of bisectors. Thus the plus sign must not be permanently identified with the internal bisector.

If the selected rays share endpoint $O$ and have unit direction vectors $\vec u,\vec v$, the internal bisector has direction $\vec u+\vec v$ and the external bisector has direction $\vec u-\vec v$. These directions identify the appropriate equation without measuring the diagram.

3. Worked examples

Example 1. Equations with normal vectors of different lengths

Consider $d_1:3x+4y-5=0$ and $d_2:8x+6y-1=0$. Their normal-vector lengths are $5$ and $10$, so

For the plus sign, $2(3x+4y-5)=8x+6y-1$ gives $2x-2y+9=0$. For the minus sign, $2(3x+4y-5)=-(8x+6y-1)$ gives $14x+14y-11=0$. The normal vectors $(2;-2)$ and $(14;14)$ have dot product $0$. The original lines meet at $O(-13/7;37/14)$, which satisfies both bisector equations.

Example 2. Choosing an angle and considering degenerate cases

For the coordinate axes $x=0$ and $y=0$, the bisectors are $y=x$ and $y=-x$. For the first-quadrant angle, the internal bisector is the ray on $y=x$; selecting a different angle may exchange these roles.

For two distinct parallel lines, the equidistant locus is the parallel line midway between them. For instance, $x=0$ and $x=4$ give $x=2$. If the lines coincide, every point of the plane is equidistant from them. The locus is the whole plane, not merely the original line, and there is no longer a unique pair of angle bisectors. In the two-sign formula, one branch may become an identity valid at every point.

4. Practice exercises

1. Find both angle-bisector lines of $x=0$ and $y=0$. Select the line containing the bisector ray of the first-quadrant angle.
2. Find the locus of points equidistant from $x=1$ and $x=5$.
3. Find the locus of points equidistant from $x+y-1=0$ and $2x+2y-2=0$. Is there a unique pair of angle-bisector lines?

5. Common mistakes

• Failing to divide each equation by the length of its normal vector.
• Using only one sign and omitting a bisector.
• Calling the plus-sign branch the internal bisector without specifying two rays.
• Applying the two-bisector conclusion to parallel or coincident lines without separate consideration.

6. Related lessons

• Distances in the plane (English lesson)
• The angle between two lines (English lesson)
• The general equation of a line (English lesson)

Original exercise resource on Scribd.