When two or more lines meet each other in a plane at a common point, they are called intersecting lines. When two lines intersect at \({90^ \circ },\) then they are called perpendicular bisectors. And a bisector divides a line into two halves. A perpendicular bisector can be drawn with a rule, a compass, and a pencil.

When two lines intersect at 90 degrees or at right angles, they are said to be perpendicular to each other. A bisector, on the other hand, is a line that divides a line into two equal halves. As a result, a perpendicular bisector of a line segment PQ denotes that it intersects PQ at 90 degrees and divides it into two equal halves. Every point in the perpendicular bisector is equidistant from the points \(P\) and \(Q.\). Let us more about how to draw a perpendicular bisector with examples

Perpendicular Definition

A line that makes an angle of \({90^ \circ }\) with another line is called perpendicular. \({90^ \circ }\) is called a right angle and is marked by a square bit between two perpendicular lines, as shown below. Here, the two lines intersect at a right angle, and hence, are said to be perpendicular to each other.

In the above figure, line \(CA\) is perpendicular to line \(AB\) and making an angle of \({90^ \circ }.\)

Bisection of a Line segment

Bisection of a line segment \(PQ\) means dividing it into two equal parts or finding a point \(O\) on \(PQ\) such that \(PO = OQ.\)

One way of bisecting a line segment is to measure it and mark off a point exactly half the length from one end of the line segment. But this method of bisecting lines is not accurate because the divisions on a ruler are limited. Therefore, an accurate measurement of the line segment is not possible.

So, let us learn the procedure of bisecting a given line segment using a ruler and compasses only. Then, the below steps are used to do the same.

Steps of Construction:

1. Draw a line segment \(PQ\) of the given length

2. With centre \(P\) and radius more than half of \(PQ,\) draw arcs, one on each side of \(PQ.\)

3. With \(Q\) as the centre and the same radius as before, draw arcs, cutting the previously drawn arcs at \(R\) and \(S\) respectively.

4. Join \(RS\) intersecting \(PQ\) at \(O.\) Then \(O\) bisects the line segment \(PQ\) as shown below.

If we measure \(PO\) and \(OQ,\) it will be equal \( \Rightarrow PO = OQ\)

Perpendicular Bisector

A line segment that intersects another line segment at a right angle and divides that line into two equal parts at its midpoint is called a perpendicular bisector.

In the above figure, \(CD\) is the perpendicular bisector of \(XY.\)

How to Draw a Perpendicular Bisector

A ruler and a compass can be used to simply draw a perpendicular bisector on a line segment. The drawn perpendicular bisector cuts the given line segment into two equal pieces exactly at its midpoint, resulting in two congruent line segments.

Proof of Perpendicular Bisector Theorem

Statement: The perpendicular bisector theorem states that any point on the perpendicular bisector is equidistant from both the endpoints of the line segment on which it is drawn.
Proof: To construct the perpendicular bisector of a given line segment \(AB\) we will follow the below construction steps.

Taking \(A\) and \(B\) as centres and radius more than \(\frac{1}{2}AB,\) draw arcs on both sides of the line segment \(AB\) (to intersect each other)
Let these arcs intersect each other at \(P\) and \(Q.\) Join \(PQ\)
Let \(PQ\) intersect \(AB\) at the point \(M.\) Then the line \(PMQ\) is the required perpendicular bisector of \(AB.\)
Now, let us see how this method gives us the perpendicular bisector of \(AB.\)

Join \(A\) and \(B\) to both \(P\) and \(Q\) to form \(AP,AQ,BP\) and \(BQ.\)
In triangles \(PAQ\) and \(PBQ,\)
\(AP = BP\) (Arcs of equal radii)
\(AQ = BQ\) (Arcs of equal radii) \(PQ = BQ\) (common)

So, according to the \(SSS\) rule, if all the three sides of one triangle are equal to the three corresponding sides of another triangle, the two triangles are congruent.
Therefore, \(\Delta PAQ \cong \Delta PBQ\)
So, \(\angle APM = \angle BPM\left({CPCT}\right)\)
Now, in \(\Delta PMA\) and \(\Delta PMB\)
\(AP = BP\) (Arcs of equal radii)
\(PM = PM\) (common)
\(\angle APM = \angle BPM\) (proved above)

So, according to the \(SAS\) rule, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle.
Therefore, \(\Delta PMA \cong \Delta PMB\)
So, \(AM = BM\) and \(\angle PMA = \angle PMB\left({CPCT} \right)\)
As, \(\angle PMA + \angle PMB = {180^ \circ }\) (Linear pair axiom)
\(\angle PMA + \angle PMB = {90^ \circ }\)
Therefore, \(PM,\) that is, \(PMQ\) is the perpendicular bisector of \(AB.\)

Perpendicular bisector Properties

1. Perpendicular bisector divides a line segment into two halves or bisects it.
2. Perpendicular bisector makes right angles with (or is perpendicular to) a line segment.
3. Every point in the perpendicular bisector is equidistant from point both the ends of a line segment.

Applications

We often use the perpendicular bisectors in geometry theorems, proofs and constructions.
1. We can use a perpendicular bisector to construct an isosceles triangle.
2. We use a perpendicular bisector to find the circumcenter of a triangle.
3. Perpendicular bisector can be used to construct a circumcircle of a triangle that passes through all three vertices of a triangle.

Solved Examples

Q.1. Draw a line segment of length \(6.6\,{\text{cm}}.\) Bisect it and measure the length of each part.
Ans: We follow the following steps of construction:
1. Draw a line segment \(AB = 6.6\,{\text{cm}}\) by using a ruler.

2. With \(A\) as centre and radius more than half of \(AB,\) draw arcs, one on each side of \(AB.\)

3. With \(B\) as the centre and the same radius as in the above step, draw arcs cutting the arcs drawn in above step at \(E\) and \(F\) respectively.

4. Draw the line segment with \(E\) and \(F\) as end-points. Suppose it meets \(AB\) at \(M.\) Then, \(M\) bisects the line segment \(AB.\)
By measuring \(AM\) and \(MB,\) we find that \(AM = MB = 3.3\,{\text{cm}}\)

Q.2. Draw a circle with the centre at point \(O\) and a radius of \(5\,{\text{cm}}.\) Draw its chord \(AB\) draw the perpendicular bisector of line segment \(AB.\) Does it pass through the centre of the circle?
Ans: Steps of construction:
With centre \(O\) and radius \(5\,{\text{cm}}\) draw a circle using the compass.

Draw a chord \(AB.\)

With \(A\) as centre and radius, more than half of \(AB\) draw arcs, one on each side of \(AB.\)

With \(B\) as the centre and the same radius as in the above step, draw arcs cutting the arcs drawn in above step at \(P\) and \(Q\) respectively.

Join \(PQ.\)

From the above figure, we can conclude that the perpendicular bisector \(PQ\) passes through the circle’s centre.

Q.3. Draw a triangle \(ABC\) with \(BC = 3.2\,{\text{cm}},\,AB = 3.6\,{\text{cm}}\) and \(\angle = {120^ \circ }.\) Also, draw a perpendicular from \(A\) on \(BC.\)
Ans: We follow the following steps to construct the required triangle.
1. Draw \(\angle XBY\) of measure \({120^ \circ }.\)
2. From ray \(BX\) cut off the line segment \(BC\) of length \(3.2\,{\text{cm}}.\)
3. From ray \(BY,\) cut off the line segment \(BA,\) of length \(3.6\,{\text{cm}}.\)
4. Join \(CA\) to obtain the required triangle.

5. Draw ray \(BZ.\)
6. With centre \(A,\) draw an arc intersecting rays \(BX\) and \(BZ\) at \(P\) and \(Q\) respectively.
7. With centre \(P\) and radius more than \(\frac{1}{2}PQ,\) cut an arc on the opposite side of \(A.\)
8. With centre \(Q\) and the same radius as the above step, cut an arc intersecting the arc drawn in the above step at \(R.\)
9. Join \(AR.\) If it meets BZ at \(L,\) then \(AL\) is the required perpendicular from \(A\) on \(BC.\)

Q.4. In a pyramid, line segment \(AD\) is the perpendicular bisector of triangle \(ABC\) to the line segment \(BC.\) If \(AB = 20\) feet and \(BD = 7\) feet, find the length of side \(AC.\)
Ans: It is given that \(AD\) is the perpendicular bisector on the line \(BC.\)
The perpendicular bisector theorem states that any point on the perpendicular bisector is equidistant from both the endpoints of the line segment on which it is drawn.
So, \(AB = AC\)
Therefore, \(AC = 20\) feet.

Q.5. Using the ruler and compass only, construct a triangle \(PQR\) in which \(QR = 6\,{\text{cm}},\angle Q = {60^ \circ }\) and \(\angle R = {75^ \circ }.\) Draw the circumcircle of the triangle.
Ans: Steps of construction:
1. Draw a line segment \(QR = 6\,{\text{cm}}\)
2. At \(Q\) draw a ray making an angle of \({60^ \circ }\) and at \(R,\) a ray making an angle of \({75^ \circ }\) which intersects each other at \(P.\)
3. Draw the perpendicular bisector of side \(QR\) and \(PR\) intersecting each other at \(O.\)
4. With centre \(O\) and radius \(OQ\) draw a circle. It will pass through \(P,Q\) and \(R.\)

Q.6. How to construct a perpendicular bisector of a triangle?
Ans: Let us consider an example to show the construction of a perpendicular bisector of a triangle.
Let us draw a perpendicular from \(A\) on \(BC\) in the triangle \(ABC\) in which, \(BC = 4\,{\text{cm}},AB = 3\,{\text{cm}}\) and \(\angle B = {45^ \circ }\)
1.Draw a line segment \(AB\) of the length of \(3\,{\text{cm}}.\)
2. Draw an angle of \({45^ \circ }\) and cut an arc at this angle at a radius of \(4\,{\text{cm}}\) at \(C.\)
3. Join \(AC\) to get the required triangle.
4. With \(A\) as a centre, draw intersecting arcs at \(M\) and \(N.\)
5. With centre \(M\) and radius more than half of \(MN,\) cut an arc on the opposite side of \(A.\)
6. With \(N\) as centre and radius, the same as in the previous step, cut an arc intersecting the last arc at \(E.\)
7. Join \(AE.\) It meets \(BC\) at \(D,\) then \(AE\) is the required perpendicular.

Summary

We have learned about the definition of perpendicular, perpendicular bisectors, construction of perpendicular bisectors, proof, properties and uses of perpendicular bisectors. Also, we have solved some problems based on the applications of perpendicular bisectors.

FAQs

Q.1. What do we call the meeting point of perpendicular bisectors of a triangle?
Ans: The meeting point of perpendicular bisectors of a triangle is called a circumcentre.

Q.2. What is a perpendicular bisector?
Ans: A line segment that intersects another line segment at a right angle and it divides that another line into two equal parts at its midpoint is called a perpendicular bisector

Q.3. What is an example of a perpendicular bisector?
Ans: The median of an equilateral triangle is an example of a perpendicular bisector.

Q.4. What is the point at which the perpendicular bisectors of a triangle meet called?
Ans: The point at which the perpendicular bisectors of a triangle meet is known as a triangle’s circumcentre.

Q.5. How to construct a perpendicular bisector of a line segment?
Ans: Steps to construct a perpendicular bisector of a line segment is as follows.
1. Draw a line segment.
2. Set compasses to longer than half the length of the line segment.
3. Construct two arcs, one centred at each end, so that two intersections are created.
4. Draw a line connecting the arc intersections.

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