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प्रश्न
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.
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उत्तर
The locus of the points inside the circle that are equidistant from the fixed points on the circle will be the diameter, which is the perpendicular bisector of the line joining the two fixed points on the circle.
संबंधित प्रश्न
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Describe the locus of vertices of all isosceles triangles having a common base.
State the locus of a point in a rhombus ABCD, which is equidistant
- from AB and AD;
- from the vertices A and C.
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Construct a rhombus ABCD with sides of length 5 cm and diagonal AC of length 6 cm. Measure ∠ ABC. Find the point R on AD such that RB = RC. Measure the length of AR.
In Δ PQR, s is a point on PR such that ∠ PQS = ∠ RQS . Prove thats is equidistant from PQ and QR.
A and B are fixed points while Pis a moving point, moving in a way that it is always equidistant from A and B. What is the locus of the path traced out by the pcint P?
Describe completely the locus of a point in the following case:
Midpoint of radii of a circle.
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.
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 :
(i) P is equidistant from AB and BC and
(ii) P is equidistant from C and D. Measure and write down the length of AB.
