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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.
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