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प्रश्न
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.
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उत्तर

- The locus of points equidistant from B and C is on BC's perpendicular bisector.
- Similarly, the locus will be at the perpendicular bisector of AB.
- The locus will be the circle that touches all three points, A, B, and C.
- The point equidistant from BA and BC will be the angle bisector of ∠ABC.
संबंधित प्रश्न
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distantce from the given line is always equal to 3 units
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Without using set squares or protractor, construct a quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6 cm, BC= 3.6 cm and CD=5 cm. Locate the point P on BD which is equidistant from BC and CD.
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Draw and describe the lorus in the following cases:
The Iocus of the mid-points of all parallel chords of a circle.
Describe completely the locus of a point in the following case:
Centre of a ball, rolling along a straight line on a level floor.
Describe completely the locus of a point in the following case:
Centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre O.
State and draw the locus of a swimmer maintaining the same distance from a lighthouse.
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.
