Use ruler and compass to answer this question. Construct ∠ABC = 90°, where AB = 6 cm and 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 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.
1. Draw the line segment BC = 8 cm.
2. At B, construct ∠ABC = 90°.
3. On the perpendicular ray, mark A such that AB = 6 cm.
(a) The locus of points equidistant from B and C is the perpendicular bisector of BC.
Construct the perpendicular bisector of BC and name it XY.
(b) The locus of points equidistant from A and B is the perpendicular bisector of AB.
Construct the perpendicular bisector of AB and name it PQ.
(c) Mark the intersection of XY and PQ as O.
A point moving at a fixed distance OA from O traces a circle.
With O as centre and OA as radius, draw a circle. This circle is the required locus.
(d) The locus of points equidistant from the intersecting lines BA and BC is their angle bisector.
Construct the angle bisector of ∠ABC and name it BZ.
Therefore:
• XY is the locus equidistant from B and C.
• PQ is the locus equidistant from A and B.
• The circle with centre O and radius OA is the fixed-distance locus.
• BZ is the locus equidistant from BA and BC.
More ICSE Hard level questions
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Prove that:
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