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

(a) Construct the locus of points equidistant from B and C.

(b) Construct the locus of points equidistant from A and B.

(c) Mark the point which satisfies both the conditions (a) and (b) as O. Construct the locus of points keeping a fixed distance OA from the fixed point O.

(d) Construct the locus of points which are equidistant from BA and BC.

Steps of construction :

1. Draw a line segment BC = 8 cm

2. Construct ∠ABC = 90°, such that AB = 6 cm.

3. Draw XY, the perpendicular bisector of BC.

4. Draw PQ, the perpendicular bisector of AB.

5. Mark point O, the intersection of segment XY and PQ.

6. Draw BZ, the angle bisector of AB and BC.

We know that,

Locus of points equidistant from two points is the perpendicular bisector of the line joining the two points segment.

(a) Locus of points equidistant from B and C is XY.

(b) Locus of points equidistant from A and B is PQ.

We know that,

Locus of points equidistant from two sides is the angular bisector of angle between them.

(d) Locus of points which are equidistant from BA and BC is BZ.