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प्रश्न
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
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उत्तर
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2
संबंधित प्रश्न
Describe the locus of a point in space, which is always at a distance of 4 cm from a fixed point.
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.
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.
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.
Construct a rhombus ABCD whose diagonals AC and BD are 8 cm and 6 cm respectively. Find by construction a point P equidistant from AB and AD and also from C and D.
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.
Describe completely the locus of a point in the following case:
Midpoint of radii of a circle.
Describe completely the locus of a point in the following case:
Point in a plane equidistant from a given line.
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.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
