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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
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
Construct a Δ XYZ in which XY= 4 cm, YZ = 5 cm and ∠ Y = 1200. Locate a point T such that ∠ YXT is a right angle and Tis equidistant from Y and Z. Measure TZ.
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 lorus in the following cases:
The locus of points at a distance of 4 cm from a fixed line.
Construct a triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.
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.
Given ∠BAC (Fig), determine the locus of a point which lies in the interior of ∠BAC and equidistant from two lines AB and AC.
Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm, BC = 8 cm.
- Construct the locus of points equidistant from B and C.
- Construct the locus of points equidistant from A and B.
- 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.
- Construct the locus of points which are equidistant from BA and BC.
