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प्रश्न
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
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उत्तर
(i) The locus of points equidistant from three given points A, B & C is the straight line PQ, which bisects AB at right angles.
(ii) Similarly, the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.
Hence, the point common to PQ and RS must satisfy both conditions; that is to say, X the point of intersection of PQ and RS will be equidistant from A, B and C.
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संबंधित प्रश्न
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
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.
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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.
Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Measure ∠ ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
In Δ PQR, bisectors of ∠ PQR and ∠ PRQ meet at I. Prove that I is equidistant from the three sides of the triangle , and PI bisects ∠ QPR .
Draw and describe the locus in the following case:
The locus of points inside a circle and equidistant from two fixed points on the circle.
Describe completely the locus of a point in the following case:
Point in a plane equidistant from a given line.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords, AB and AC, of the circle of length 6 cm and 5 cm respectively.
- Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
- Construct the locus of points, inside the circle, that are equidistant from AB and AC.
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.
