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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.
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संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
Use ruler and compasses only for this question.
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