Advertisements
Advertisements
Question
In the figure given below, find a point P on CD equidistant from points A and B.

Advertisements
Solution

Steps of construction:
- AB and CD are the two lines given.
- Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.
APPEARS IN
RELATED QUESTIONS
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.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
Describe the locus of points at distances greater than or equal to 35 mm from a given point.
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.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
