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प्रश्न
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
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उत्तर
(i) The locus of points equidistant from three given points A, B & C is the straight line PQ, which bisects AB at right angles.
(ii) Similarly, the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.
Hence, the point common to PQ and RS must satisfy both conditions; that is to say, X the point of intersection of PQ and RS will be equidistant from A, B and C.
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संबंधित प्रश्न
On a graph paper, draw the lines x = 3 and y = –5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Describe the locus of a point in space, which is always at a distance of 4 cm from a fixed point.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
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.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. prove your construction.
(ii) Construct the locus of points, inside the circle that are equidistant from AB and AC.
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.
Construct a triangle BPC given BC = 5 cm, BP = 4 cm and .
i) complete the rectangle ABCD such that:
a) P is equidistant from AB and BCV
b) P is equidistant from C and D.
ii) Measure and record the length of AB.
State and draw the locus of a point equidistant from two given parallel lines.
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.
Without using set squares or a protractor, construct:
- Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
- Draw the locus of a point which moves so that it is always 2.5 cm from B.
- Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
- Mark the point of intersection of the loci with the letter P and measure PC.
