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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.
RELATED QUESTIONS
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 vertices of all isosceles triangles having a common base.
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AB2 = AP2 + BP2,
where A and B are two fixed points.
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Using only a ruler and compass construct ∠ABC = 120°, where AB = BC = 5 cm.
(i) Mark two points D and E which satisfy the condition that they are equidistant from both ABA and BC.
(ii) In the above figure, join AD, DC, AE and EC. Describe the figures:
(a) AECB, (b) ABD, (c) ABE.
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.
