Advertisements
Advertisements
प्रश्न
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Advertisements
उत्तर
Steps of construction:
- Draw a line segment AB of 6 cm.
- Draw perpendicular bisector LM of AB. LM is the required locus.
- Take any point on LM say P.
- Join PA and PB. Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
APPEARS IN
संबंधित प्रश्न
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.
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 points at a distance 2 cm from a fixed line.
Describe the locus of a stone dropped from the top of a tower.
The speed of sound is 332 metres per second. A gun is fired. Describe the locus of all the people on the earth’s surface, who hear the sound exactly one second later.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
Find the locus of points which are equidistant from three non-collinear points.
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
