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Question
Find the locus of points which are equidistant from three non-collinear points.
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Solution
Let A, B and C be three non-collinear points. Join AB and BC. Let P be a moving point. Since, P is equidistant from A and B, it follows that P lies on the perpendicular bisector of AB.
Again P is equidistant from B and C. So, P lies on the perpendicular bisector of BC.
Thus, P is the point of intersection of the perpendicular bisector of AB and BC. So, P coincides three given non-collinear points. Hence, the required locus is the centre of the circle passing through three given non-collinear points.
RELATED QUESTIONS
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
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Use ruler and compasses only for this question.
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