Advertisements
Advertisements
Question
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.
Advertisements
Solution
Construction: Join PQ which meets AB in D.


Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.
APPEARS IN
RELATED QUESTIONS
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
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 a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
Describe the locus of points at distances less than 3 cm from a given point.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
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.


