Construction: Perpendicular Bisector
A perpendicular bisector of a line segment is a line that passes through the midpoint of the segment and is perpendicular (at 90°) to it.
In Class 9 Constructions, students learn to construct the perpendicular bisector of a given line segment using only a compass and a straightedge (ruler without markings). The construction must be accompanied by a justification using geometric reasoning.
The perpendicular bisector has a key property: every point on it is equidistant from the two endpoints of the line segment.
What is Construction: Perpendicular Bisector?
Definition: The perpendicular bisector of a line segment AB is a line that:
- Passes through the midpoint M of AB (so AM = MB)
- Is perpendicular to AB (makes a 90° angle with AB)
Locus Property:
Every point on the perpendicular bisector of AB is equidistant from A and B.
If P lies on the perpendicular bisector of AB, then PA = PB
Construction: Perpendicular Bisector Formula
Key Properties:
- The perpendicular bisector divides the segment into two equal halves.
- It makes a 90° angle with the segment.
- Any point on it is equidistant from both endpoints.
- The circumcentre of a triangle is the intersection of the perpendicular bisectors of all three sides.
In coordinate geometry:
- Midpoint of A(x₁, y₁) and B(x₂, y₂) is M = ((x₁+x₂)/2, (y₁+y₂)/2)
- Slope of AB = (y₂ − y₁)/(x₂ − x₁)
- Slope of perpendicular bisector = −1 / slope of AB
Derivation and Proof
Steps of Construction:
Given: A line segment AB.
Required: Construct the perpendicular bisector of AB.
Steps:
- Draw the given line segment AB.
- With A as centre, take a radius greater than ½AB and draw arcs on both sides of AB.
- With B as centre and the same radius, draw arcs on both sides of AB, cutting the previous arcs at points P and Q.
- Join P and Q. The line PQ is the perpendicular bisector of AB.
- PQ intersects AB at the midpoint M.
Justification:
- By construction, AP = BP (equal radii from step 2 and 3).
- Also, AQ = BQ (equal radii from step 2 and 3).
- Therefore, P and Q are both equidistant from A and B.
- The locus of points equidistant from A and B is the perpendicular bisector of AB.
- Since both P and Q lie on this locus, line PQ is the perpendicular bisector of AB.
Alternatively, using congruent triangles:
- In ▵APM and ▵BPM: AP = BP (by construction), AM = BM (M is midpoint), PM = PM (common).
- By SSS, ▵APM ≅ ▵BPM.
- Therefore ∠PMA = ∠PMB. Since they form a linear pair, each = 90°.
- Hence PQ ⊥ AB and M is the midpoint. PQ is the perpendicular bisector.
Types and Properties
Related Constructions:
1. Finding the Midpoint of a Segment
- The perpendicular bisector construction automatically gives the midpoint.
- M, the point where PQ crosses AB, is the midpoint of AB.
2. Constructing a 90° Angle at a Point on a Line
- The perpendicular bisector method can be adapted to construct a right angle at any point on a line.
3. Circumcentre of a Triangle
- Construct the perpendicular bisectors of any two sides of a triangle.
- Their intersection is the circumcentre, equidistant from all three vertices.
4. Dropping a Perpendicular from an External Point
- A related construction, but uses arcs from a point outside the line to find the foot of the perpendicular.
Solved Examples
Example 1: Example 1: Construct perpendicular bisector of a 6 cm segment
Problem: Draw a line segment AB = 6 cm and construct its perpendicular bisector.
Solution:
- Draw AB = 6 cm.
- With A as centre and radius 4 cm (more than 3 cm = ½ × 6), draw arcs above and below AB.
- With B as centre and same radius 4 cm, draw arcs cutting the previous arcs at P (above) and Q (below).
- Join PQ. It crosses AB at M.
Verification:
- Measure AM and MB. Both = 3 cm. ✓
- Measure ∠PMA. It is 90°. ✓
Answer: PQ is the perpendicular bisector of AB, and M is the midpoint.
Example 2: Example 2: Verify the equidistant property
Problem: After constructing the perpendicular bisector of AB = 8 cm, take any point P on the bisector. Verify that PA = PB.
Solution:
- Construct AB = 8 cm and its perpendicular bisector PQ.
- Pick any point R on line PQ.
- Measure RA and RB using a compass or ruler.
Result:
- RA = RB for every point R on PQ.
- This verifies the locus property of the perpendicular bisector.
Answer: Every point on the perpendicular bisector is equidistant from A and B. Verified.
Example 3: Example 3: Construct the circumcentre of a triangle
Problem: Construct the circumcentre of ▵ABC with AB = 6 cm, BC = 7 cm, AC = 8 cm.
Solution:
- Draw ▵ABC with the given side lengths.
- Construct the perpendicular bisector of AB.
- Construct the perpendicular bisector of BC.
- Mark their intersection point as O (circumcentre).
- Measure OA, OB, OC — all three are equal.
Answer: O is the circumcentre. OA = OB = OC = circumradius.
Example 4: Example 4: Locate the midpoint by construction
Problem: Without using a ruler to measure, find the midpoint of a line segment PQ = 10 cm.
Solution:
- Draw PQ = 10 cm.
- With P as centre, radius greater than 5 cm (say 7 cm), draw arcs above and below PQ.
- With Q as centre and same radius, draw arcs intersecting the first at A and B.
- Join AB. Let it intersect PQ at M.
- M is the midpoint of PQ.
Verification:
- PM = MQ = 5 cm (measure to verify)
Answer: M is the midpoint of PQ.
Example 5: Example 5: Construction with justification
Problem: Construct the perpendicular bisector of AB = 5.4 cm and justify using congruent triangles.
Solution:
Construction:
- Draw AB = 5.4 cm.
- With A and B as centres, radius = 4 cm, draw arcs meeting at P and Q.
- Join PQ, meeting AB at M.
Justification:
- AP = BP = 4 cm (by construction).
- AQ = BQ = 4 cm (by construction).
- PQ = PQ (common side).
- By SSS, ▵APQ ≅ ▵BPQ.
- So ∠APM = ∠BPM (CPCT).
- In ▵APM and ▵BPM: AP = BP, PM = PM (common), ∠APM = ∠BPM.
- By SAS, ▵APM ≅ ▵BPM.
- So AM = BM (CPCT) and ∠PMA = ∠PMB = 90°.
Answer: PQ is the perpendicular bisector of AB. Justified.
Example 6: Example 6: Error analysis
Problem: A student used radius = 2 cm to construct the perpendicular bisector of a segment of length 7 cm. Why did the construction fail?
Solution:
Analysis:
- Half the segment = 3.5 cm.
- The chosen radius (2 cm) is less than 3.5 cm.
- The arcs from A and B do not intersect because the circles are too small to meet.
Rule:
- The radius must be greater than half the length of the segment.
- Here, radius must be > 3.5 cm.
Answer: The construction failed because 2 cm < 3.5 cm (½ of 7 cm). The arcs did not intersect.
Example 7: Example 7: Perpendicular bisector in a real-world context
Problem: Two villages A and B are 10 km apart. A hospital must be built equidistant from both. Describe the possible locations.
Solution:
Given:
- A and B are 10 km apart.
- The hospital must satisfy PA = PB.
Using the perpendicular bisector property:
- The set of all points equidistant from A and B is the perpendicular bisector of AB.
- Any point on this line is exactly 5 km or more from A and B.
- The closest point to both is the midpoint M, which is 5 km from each village.
Answer: The hospital can be built at any point on the perpendicular bisector of the line joining the two villages.
Example 8: Example 8: Constructing a right angle
Problem: Using the perpendicular bisector method, construct a 90° angle at the midpoint of AB = 8 cm.
Solution:
- Draw AB = 8 cm.
- With A as centre, radius 5 cm, draw arcs above and below AB.
- With B as centre and radius 5 cm, draw arcs meeting at P and Q.
- Join PQ. It meets AB at M (midpoint).
- ∠PMA = 90°.
Verification:
- AM = MB = 4 cm ✓
- ∠PMA = 90° ✓
Answer: A 90° angle is constructed at M on line AB.
Real-World Applications
Applications:
- Finding Midpoints: The perpendicular bisector construction is the standard method to bisect a line segment without measurement.
- Circumcentre: Intersection of perpendicular bisectors of a triangle's sides gives the circumcentre, the centre of the circumscribed circle.
- Equal Distance Problems: Locating a point equidistant from two given points (hospital, tower, well placement problems).
- Mirror Lines in Symmetry: The perpendicular bisector of a line joining a point to its reflection is the mirror line (axis of symmetry).
- Engineering and Architecture: Used in layout planning, road construction midpoints, and structural alignment.
Key Points to Remember
- The perpendicular bisector of a segment passes through its midpoint at 90°.
- Construction requires a compass and straightedge only.
- The radius used must be greater than half the length of the segment.
- The same radius must be used from both endpoints.
- Every point on the perpendicular bisector is equidistant from both endpoints.
- The perpendicular bisector also gives the midpoint of the segment.
- The circumcentre of a triangle is found by intersecting perpendicular bisectors of its sides.
- Justification can use the SSS or SAS congruence rule.
- This construction is a prerequisite for more advanced constructions in Class 9.
- In NCERT Class 9, all constructions must be accompanied by a written justification.
Practice Problems
- Draw a line segment of length 7.5 cm and construct its perpendicular bisector. Verify by measurement.
- Construct the perpendicular bisector of PQ = 9 cm. Mark three points on the bisector and verify that each is equidistant from P and Q.
- Draw a triangle with sides 5 cm, 6 cm, and 7 cm. Construct the perpendicular bisectors of any two sides to find the circumcentre.
- Using only a compass and straightedge, find the midpoint of a segment AB = 11 cm.
- Construct the perpendicular bisector of a segment of length 4.8 cm. Write the justification using congruent triangles.
- Two points X and Y are 6 cm apart. Construct the locus of all points equidistant from X and Y.
Frequently Asked Questions
Q1. What is a perpendicular bisector?
A perpendicular bisector of a line segment is a line that passes through the midpoint of the segment and is perpendicular (at 90°) to it.
Q2. Why must the compass radius be more than half the segment length?
If the radius is less than or equal to half the length, the arcs drawn from the two endpoints will not intersect. The arcs must overlap to create intersection points P and Q.
Q3. What is the equidistant property of the perpendicular bisector?
Every point on the perpendicular bisector of AB is at equal distance from A and B. Conversely, any point equidistant from A and B lies on the perpendicular bisector.
Q4. How is the perpendicular bisector related to the circumcentre?
The circumcentre of a triangle is the point where the perpendicular bisectors of all three sides meet. It is equidistant from all three vertices and is the centre of the circumscribed circle.
Q5. Is this construction in the CBSE Class 9 syllabus?
Yes. Perpendicular bisector construction with justification is part of Chapter 11 (Constructions) of NCERT Class 9 Mathematics.
Q6. Why do we need to justify the construction?
In Class 9, NCERT requires students to not only perform the construction but also prove that the construction is correct using geometric reasoning such as congruence rules.
Q7. Can the perpendicular bisector be constructed for a curved line?
No. The perpendicular bisector is defined only for a straight line segment. For curves, the concept of perpendicular bisector does not directly apply.
Q8. What is the difference between a perpendicular bisector and an angle bisector?
A perpendicular bisector bisects a line segment at 90°. An angle bisector divides an angle into two equal parts. They are different constructions with different purposes.
