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प्रश्न
A and B are fixed points while Pis a moving point, moving in a way that it is always equidistant from A and B. What is the locus of the path traced out by the pcint P?
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उत्तर
The locus of path traced by point P equidistant from A and B is the perpendicular bisector of the line segment joining the two points.
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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:
- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
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.
Ruler and compasses 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.
- Construct a ΔABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
- Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
- Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
- Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
- Measure and record the length of CQ.
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
In Δ ABC, B and Care fixed points. Find the locus of point A which moves such that the area of Δ ABC remains the same.
Using only ruler and compasses, construct a triangle ABC 1 with AB = 5 cm, BC = 3.5 cm and AC= 4 cm. Mark a point P, which is equidistant from AB, BC and also from Band C. Measure the length of PB.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.
Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
(ii) In the above figure, join AD, DC, AE and EC. Describe the figures:
(a) AECB, (b) ABD, (c) ABE.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
