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प्रश्न
Find the locus of points which are equidistant from three non-collinear points.
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उत्तर
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.
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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.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from A and B.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Describe the locus of points at distances less than 3 cm from a given point.
Describe the locus of points at distances greater than 4 cm from a given point.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
- the locus of the centres of all circles which touch AB and AC,
- the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC .
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
