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प्रश्न
Describe completely the locus of a point in the following case:
Point in a plane equidistant from a given line.
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उत्तर
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.
संबंधित प्रश्न
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 a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
- Plot the points A(1, 1), B(5, 3) and C(2, 7).
- Construct the locus of points equidistant from A and B.
- Construct the locus of points equidistant from AB and AC.
- Locate the point P such that PA = PB and P is equidistant from AB and AC.
- Measure and record the length PA in cm.
Draw and describe the locus in the following case:
The locus of points inside a circle and equidistant from two fixed points on the circle.
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.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
State and draw the locus of a point equidistant from two given parallel lines.
Without using set squares or protractor.
(i) Construct a ΔABC, given BC = 4 cm, angle B = 75° and CA = 6 cm.
(ii) Find the point P such that PB = PC and P is equidistant from the side BC and BA. Measure AP.
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.
