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प्रश्न
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
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उत्तर
Steps of construction:
- Draw a line segment AB of 6 cm.
- Draw perpendicular bisector LM of AB. LM is the required locus.
- Take any point on LM say P.
- Join PA and PB. Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
संबंधित प्रश्न
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
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