Advertisements
Advertisements
प्रश्न
Find the locus of points which are equidistant from three non-collinear points.
Advertisements
उत्तर
Let A, B and C be three non-collinear points. Join AB and BC. Let P be a moving point. Since, P is equidistant from A and B, it follows that P lies on the perpendicular bisector of AB.
Again P is equidistant from B and C. So, P lies on the perpendicular bisector of BC.
Thus, P is the point of intersection of the perpendicular bisector of AB and BC. So, P coincides three given non-collinear points. Hence, the required locus is the centre of the circle passing through three given non-collinear points.
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.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from A and B.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
- the locus of the centres of all circles which touch AB and AC,
- the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC .
In Δ ABC, the perpendicular bisector of AB and AC meet at 0. Prove that O is equidistant from the three vertices. Also, prove that if M is the mid-point of BC then OM meets BC at right angles.
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.
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.
