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Practical Geometry - Constructing a Perpendicular Bisector of a Line Segment

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A line segment is a fixed portion of a line with two distinct endpoints, denoted as AB‾\overline{AB}. Visually, it is represented as a straight path with a starting point and an ending point.

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A bisector is a line, ray, or segment that divides a line segment into two equal parts. Visually, if you fold a paper so that the two endpoints of a line segment meet, the crease formed is the bisector.

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Perpendicularity occurs when two lines or segments intersect at a right angle (90∘90^{\circ}). Visually, this looks like the meeting of a horizontal floor and a vertical wall, or the corner of a perfect square.

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A Perpendicular Bisector is a line that is both perpendicular to a line segment and passes through its midpoint. It divides the segment into two equal halves while forming 90∘90^{\circ} angles at the point of intersection.

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To construct a perpendicular bisector using a compass, the radius of the compass must be set to more than half the length of the segment (i.e., r>12×lengthr > \frac{1}{2} \times \text{length}). If the radius is too small, the arcs drawn from the endpoints will not intersect.

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The construction process involves drawing two arcs from each endpoint (one above and one below the segment). These four arcs intersect at two distinct points. Visually, these look like two 'X' shapes above and below the line segment.

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The point where the perpendicular bisector crosses the segment is called the midpoint MM. This point MM is equidistant from both endpoints AA and BB, meaning AM=MBAM = MB.

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Any point PP lying on the perpendicular bisector is equidistant from the endpoints of the segment. This means that if you draw lines from PP to endpoints AA and BB, then PA=PBPA = PB.

📐Formulae

Length of each bisected part: AM=MB=12×ABAM = MB = \frac{1}{2} \times AB

Condition for arc intersection: r>AB2r > \frac{AB}{2} (where rr is the compass radius)

Angle of intersection: ∠PMA=∠PMB=90∘\angle PMA = \angle PMB = 90^{\circ}

Total length: AB=AM+MBAB = AM + MB

💡Examples

Problem 1:

Draw a line segment AB‾\overline{AB} of length 7 cm7 \text{ cm} and construct its perpendicular bisector using a ruler and compass.

Solution:

  1. Draw a line segment AB‾=7 cm\overline{AB} = 7 \text{ cm} using a ruler.
  2. With AA as the center and a radius more than half of ABAB (e.g., 4 cm4 \text{ cm}), draw two arcs—one above AB‾\overline{AB} and one below it.
  3. Keeping the same radius and with BB as the center, draw two more arcs intersecting the previous arcs at points PP and QQ.
  4. Join point PP to point QQ using a ruler. Let PQPQ intersect ABAB at point MM.
  5. The line PQPQ is the required perpendicular bisector, and MM is the midpoint.
  6. Verification: Measure AMAM and MBMB with a ruler. Both should be 3.5 cm3.5 \text{ cm}, since 7 cm2=3.5 cm\frac{7 \text{ cm}}{2} = 3.5 \text{ cm}.

Explanation:

The compass radius must be greater than 3.5 cm3.5 \text{ cm} to ensure the arcs from AA and BB cross each other. The points of intersection PP and QQ provide the vertical path that cuts AB‾\overline{AB} exactly in half at a 90∘90^{\circ} angle.

Problem 2:

If a line XYXY is the perpendicular bisector of a segment PQ‾\overline{PQ} and they intersect at point OO, find the length of PQPQ if PO=4.2 cmPO = 4.2 \text{ cm}.

Solution:

  1. Since XYXY is the perpendicular bisector of PQ‾\overline{PQ}, point OO must be the midpoint of PQ‾\overline{PQ}.
  2. By the property of a midpoint, PO=OQPO = OQ.
  3. Given PO=4.2 cmPO = 4.2 \text{ cm}, it follows that OQ=4.2 cmOQ = 4.2 \text{ cm}.
  4. The total length PQ=PO+OQPQ = PO + OQ.
  5. PQ=4.2 cm+4.2 cm=8.4 cmPQ = 4.2 \text{ cm} + 4.2 \text{ cm} = 8.4 \text{ cm}.

Explanation:

This problem uses the definition of a bisector. A perpendicular bisector always passes through the midpoint, meaning it splits the segment into two equal halves. Multiplying the length of one half by 22 gives the total length.