Question

Use a graph paper for this question. Take 1 cm = 1 unit on both the axes.

Plot the points A(2,2), B(6,4) and C(3,8). Construct the locus of points equidistant from A and B. Also, construct the locus of points equidistant from AB and AC. Locate the point P such that PA = PB and P is equidistant from AB and AC. Then the length of PA in cm is

• A. 2
• B. 2.5
• C. 2.6
• D. 3

Answer 1

The correct option is B

2.5

Steps of construction:

• Plot the points A(2,2), B(6,4) and C(3,8) on a graph and join AB, BC and CA.
• Draw the perpendicular bisector 'l' of AB and angle bisector 'm' of angle A which intersect each other at P.

We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points. Now, since P lies on the perpendicular bisector of AB, P is equidistant from A and B. Then PA = PB.

Again, we know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines. Now, since P lies on the angle bisector of angle A, P is equidistant from AB and AC.

Hence P is the required point.

On measuring, we get, PA = 2.5 cm