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प्रश्न
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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उत्तर
Steps of oonstruction:
(i) Draw a line segment AB of 9 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
Proof:
(i) Take any point on LM say P.
(ii) Join PA and PB.
Since, Plies on the right bisector of line AB.
Therefore, Pis 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
Describe the locus of vertices of all isosceles triangles having a common base.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
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Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
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Ruler and compass only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
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