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प्रश्न
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
- always 4 cm from the line AB.
- equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
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उत्तर
Steps of construction:
i. Draw a line segment AB equal to 8 cm.
ii. Draw two parallel lines l and m to AB at a distance of 4 cm.
iii. Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
iv. Join AX, AY, BX and BY.
The figure so formed is a square as its diagonals are equal and intersect at 90°.
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संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:
- Point A is equidistant from all the three sides of the triangle.
- AM bisects angle LMN.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of a stone dropped from the top of a tower.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.
