Locus of a Point Equidistant from Two Intersecting Lines

Q.

Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and $\angle BAC={105}^{\circ }$. Mark the point of intersection of the locus of points equidistant from BA and BC and the locus of points equidistant from B and C, as P. Then the length of PC is

1. 4 cm

2. 4.2 cm

3. 4.8 cm

4. 5.2 cm

Q.

In triangle ABC, the construction required to find point P equidistant from AB and AC and also, equidistant from B and C is/are

1. Bisector of angle A
   

2. Perpendicular bisector of side BC
   

3. Circle taking BC as the diameter
   

4. None of these

Q.

The bisectors of angle A and C of a quadrilateral ABCD intersect each other at P. Then P is equidistant from the opposite sides AB and CD.

1. True

2. False