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प्रश्न
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?
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उत्तर
The locus of path traced by point P equidistant from A and B is the perpendicular bisector of the line segment joining the two points.
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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.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Plot the points A(2, 9), B(–1, 3) and C(6, 3) on graph paper. On the same graph paper draw the locus of point A so that the area of ΔABC remains the same as A moves.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
Without using set squares or protractor, construct a quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6 cm, BC= 3.6 cm and CD=5 cm. Locate the point P on BD which is equidistant from BC and CD.
Construct a ti.PQR, in which PQ=S. 5 cm, QR=3. 2 cm and PR=4.8 cm. Draw the locus of a point which moves so that it is always 2.5 cm from Q.
Construct a Δ XYZ in which XY= 4 cm, YZ = 5 cm and ∠ Y = 1200. Locate a point T such that ∠ YXT is a right angle and Tis equidistant from Y and Z. Measure TZ.
Using only ruler and compasses, construct a triangle ABC 1 with AB = 5 cm, BC = 3.5 cm and AC= 4 cm. Mark a point P, which is equidistant from AB, BC and also from Band C. Measure the length of PB.
Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.
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.
