English

A and B Are Fixed Points While Pis a Moving Point, Moving in a Way that It is Always Equidistant from a and B

Advertisements
Advertisements

Question

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? 

Diagram
Advertisements

Solution

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. 

shaalaa.com
  Is there an error in this question or solution?
Chapter 15: Loci - Exercise 16.1

APPEARS IN

Frank Mathematics Part 2 [English] Class 10 ICSE
Chapter 15 Loci
Exercise 16.1 | Q 12

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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.

  1. Construct a ΔABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
  2. Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
  3. Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
  4. Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
  5. Measure and record the length of CQ.

Use graph paper for this question. Take 2 cm = 1 unit on both the axes.

  1. Plot the points A(1, 1), B(5, 3) and C(2, 7).
  2. Construct the locus of points equidistant from A and B.
  3. Construct the locus of points equidistant from AB and AC.
  4. Locate the point P such that PA = PB and P is equidistant from AB and AC.
  5. Measure and record the length PA in cm. 

Two straight roads AB and CD cross each other at Pat an angle of 75°  . X is a stone on the road AB, 800m from P towards B. BY taking an appropriate scale draw a figure to locate the position of a pole, which is equidistant from P and X, and is also equidistant from the roads. 


Describe completely the locus of a point in the following case:

Point in a plane equidistant from a given line. 


Use ruler and compass only for the following question. All construction lines and arcs must be clearly shown.

  1. Construct a ΔABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
  2. Construct the locus of points at a distance of 3.5 cm from A.
  3. Construct the locus of points equidistant from AC and BC.
  4. Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.

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 ruler and compasses construct:
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of point equidistant from A and C.
(iii) a circle touching AB at A and passing through C.


How will you find a point equidistant from three given points A, B, C which are not in the same straight line?


Without using set squares or protractor.
(i) Construct a ΔABC, given BC = 4 cm, angle B = 75° and CA = 6 cm.
(ii) Find the point P such that PB = PC and P is equidistant from the side BC and BA. Measure AP.


Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.

  1. Construct the locus of points equidistant from B and C.
  2. Construct the locus of points equidistant from A and B.
  3. Mark the point which satisfies both the conditions (a) and (b) as 0. Construct the locus of points keeping a fixed distance OA from the fixed point 0.
  4. Construct the locus of points which are equidistant from BA and BC.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×