Parallelism and perpendicularity are two properties of straight lines which are discussed in detail in geometry. These two special attributes make other geometrical concepts pronounced. Bisector is a line which divides a line or an angle into two equal halves. When we consider a line segment $\overline{AB}$, then the bisection can be achieved by drawing a line through the mid point of AB. Since infinite number of lines can be drawn through a given point, the line segment can have many bisectors, all lines passing through the mid point of $\overline{AB}$, perpendicular bisector is a line which passes through midpoint AB and also perpendicular to it.

### What is a Perpendicular Bisector?

Perpendicular bisector of a line segment is a line that bisects it at right angles.

The diagram below shows the different bisectors to the line segment $\overline{AB}$. All the lines passing through the midpoint C are bisectors to $\overline{AB}$. While the other bisectors B_{1}, B_{2} do not intersect AB at 90^{o}, PB the perpendicular bisector of the line segment $\overline{AB}$ bisects $\overline{AB}$ at right angles.

Perpendicular bisector of a line segment is a line that bisects it at right angles.

The diagram below shows the different bisectors to the line segment $\overline{AB}$. All the lines passing through the midpoint C are bisectors to $\overline{AB}$. While the other bisectors B

In analytical geometry analytical geometry, if the coordinates of the points A and B are known then the equation to the perpendicular bisector of the line joining the points A and B can be determined.

Let the coordinates of the points A and B be (x_{1}, y_{1}) and (x_{2}, y_{2}).

Hence the slope of the line joining A and B = $\frac{y_2-y_1}{x_2-x_1}$.

The slope of the line perpendicular to AB is the negative reciprocal of this.

Slope of the perpendicular bisector to AB = $\frac{y_2-y_1}{x_2-x_1}$.

The perpendicular bisector of AB passes through the midpoint AB $\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$.

Using the point slope formula, the equation of the perpendicular bisector can be written as,

y-$\frac{y_1+y_2}{2}$ = $\frac{x_1-x_2}{y_2-y_1},(x\frac{x_1+x_2}{2})$

simplifying taking 2 (y_{2 }- y_{1}), as the LCD,

2y(y_{2 }- y_{1}) - (y_{1 }+ y_{2})(y_{2 }- y_{1}) = (x_{1 }- x_{2})(2x-(x_{1 }+ x_{2}))

2x(x_{2}-x_{1}) + 2y(y_{2}-y_{1}) = x^{2}_{2 }- x^{2}_{1 }+ y^{2}_{2 }- y^{2}_{1}

### Perpendicular Bisector Equation

The equation to the perpendicular bisector can be written in general form of equation to a line as follows:

**2x(x**_{2 }- x_{1}) + 2y(y_{2 }- y_{1}) = $\frac{x_2 ^2-x_1 ^2+y_2 ^2-y_1 ^2}{2}$### How to Construct a Perpendicular Bisector

Let the coordinates of the points A and B be (x

Hence the slope of the line joining A and B = $\frac{y_2-y_1}{x_2-x_1}$.

The slope of the line perpendicular to AB is the negative reciprocal of this.

Slope of the perpendicular bisector to AB = $\frac{y_2-y_1}{x_2-x_1}$.

The perpendicular bisector of AB passes through the midpoint AB $\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$.

Using the point slope formula, the equation of the perpendicular bisector can be written as,

y-$\frac{y_1+y_2}{2}$ = $\frac{x_1-x_2}{y_2-y_1},(x\frac{x_1+x_2}{2})$

simplifying taking 2 (y

2y(y

2x(x

The points of concurrency like circumference, orthocenter, incenter and centroid are important aspects of triangles. The perpendicular bisectors of the triangle form one of these points.

The perpendicular bisectors of the sides of the triangle are concurrent. The point of concurrency is called the circumcenter of the triangle.

In the adjoining diagram, the perpendicular bisectors of sides AB, BC and CA meet at point O.

Since O is a point on the perpendicular bisector of AB, it is equidistant from A and B. By similar arguments it is also equidistant from B and C as well. This makes the point O equidistant from all the three vertices of the triangle A,B and C.

Hence the circle drawn with O as center and OA as radius will pass through all the three vertices A, B and C. This circle is called the circumcircle of the triangle and hence the point O is known as the circumcenter of the triangle ABC.

In the case of an equilateral triangle all the four points of concurrency the circumcenter, orthocenter, incenter and centroid are coincident.

**Perpendicular bisectors in quadrilaterals**

The diagonals of a rhombus or a square are perpendicular bisectors of each other. In the case of a kite the longer diagonal is the perpendicular bisector of the shorter.

**Perpendicular bisectors in circles**

The center of a circle is the point concurrence of perpendicular bisectors of all chords of the circle. For a given chord the center of the circle lie on the perpendicular bisector of the chord, In this context the perpendicular bisector of a line segment can also be viewed as the locus of centers of all circles for which the line segment is a chord.

Two mutually perpendicular diameters of a circle bisect each other.

The perpendicular bisectors of the sides of the triangle are concurrent. The point of concurrency is called the circumcenter of the triangle.

In the adjoining diagram, the perpendicular bisectors of sides AB, BC and CA meet at point O.

Since O is a point on the perpendicular bisector of AB, it is equidistant from A and B. By similar arguments it is also equidistant from B and C as well. This makes the point O equidistant from all the three vertices of the triangle A,B and C.

Hence the circle drawn with O as center and OA as radius will pass through all the three vertices A, B and C. This circle is called the circumcircle of the triangle and hence the point O is known as the circumcenter of the triangle ABC.

In the case of an equilateral triangle all the four points of concurrency the circumcenter, orthocenter, incenter and centroid are coincident.

The diagonals of a rhombus or a square are perpendicular bisectors of each other. In the case of a kite the longer diagonal is the perpendicular bisector of the shorter.

The center of a circle is the point concurrence of perpendicular bisectors of all chords of the circle. For a given chord the center of the circle lie on the perpendicular bisector of the chord, In this context the perpendicular bisector of a line segment can also be viewed as the locus of centers of all circles for which the line segment is a chord.

Two mutually perpendicular diameters of a circle bisect each other.

We know that C the midpoint of $\overline{AB}$, is equidistant from A and B. Is C the only such point? It can be proved that using triangle congruence that any point equidistant from the two points lie on the perpendicular bisector of the line joining the two points. Conversely any point on the perpendicular bisector of AB is equidistance from A and B.

Is equidistant from A and B. Is C the only such point? It can be proved that using triangle congruence that any point equidistant from the two points lie on the perpendicular bisector of the line joining the two points. Conversely any point on the perpendicular bisector of AB is equidistance from A and B.

In general an arbitrary point 'P' on the perpendicular bisector of line segment $\overline{AB}$ is equidistant from the end points A and B of the segment.

' M ' the midpoint of line segment $\overline{AB}$ is the point of intersection of AB and its perpendicular bisector. Hence M is also a point on the perpendicular bisector.

Based on the above fact, the perpendicular bisector theorem and its converse can be stated as follows:

### Perpendicular Bisector Theorem

If P is a point equidistant from A and B, then P lies on the perpendicular bisector of $\overline{AB}$.

**Converse of perpendicular theorem**

If PM is the Perpendicular bisector of $\overline{AB}$ , then P is equidistant from A and B.

Since the theorem is true both ways.

**Perpendicular bisector of a line segment is the locus of all points that are equidistant from the end points.**

Construction of Perpendicular bisector: This locus property of the perpendicular bisector gives a method for constructing a perpendicular bisector for a given line segment.

Keeping the same radius and the points as A and B as centers draw arcs both above and below the line segment AB. Let the arcs cut each other at P and Q as shown in the diagram. Join PQ, PQ is the perpendicular bisector of the line segment$\overline{AB}$. The perpendicular bisector is constructed to determne the midpoint of a line segment.

Is equidistant from A and B. Is C the only such point? It can be proved that using triangle congruence that any point equidistant from the two points lie on the perpendicular bisector of the line joining the two points. Conversely any point on the perpendicular bisector of AB is equidistance from A and B.

In general an arbitrary point 'P' on the perpendicular bisector of line segment $\overline{AB}$ is equidistant from the end points A and B of the segment.

' M ' the midpoint of line segment $\overline{AB}$ is the point of intersection of AB and its perpendicular bisector. Hence M is also a point on the perpendicular bisector.

Based on the above fact, the perpendicular bisector theorem and its converse can be stated as follows:

If P is a point equidistant from A and B, then P lies on the perpendicular bisector of $\overline{AB}$.

If PM is the Perpendicular bisector of $\overline{AB}$ , then P is equidistant from A and B.

Since the theorem is true both ways.

Construction of Perpendicular bisector: This locus property of the perpendicular bisector gives a method for constructing a perpendicular bisector for a given line segment.

Keeping the same radius and the points as A and B as centers draw arcs both above and below the line segment AB. Let the arcs cut each other at P and Q as shown in the diagram. Join PQ, PQ is the perpendicular bisector of the line segment$\overline{AB}$. The perpendicular bisector is constructed to determne the midpoint of a line segment.