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Question
Describe completely the locus of a point in the following case:
Point in a plane equidistant from a given line.
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Solution
The locus of all points in a plane equidistant from a fixed line is represented by two parallel lines on either side of it at a distance away.
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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.
Use ruler and compasses only for the following questions:
Construct triangle BCP, when CB = 5 cm, BP = 4 cm, ∠PBC = 45°.
Complete the rectangle ABCD such that :
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Ruler and compass only 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.
(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point 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 as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
