Question

Use graph paper for this question. Take 2 cm = 1 unit on both the axes.

1. Plot the points A(1, 1), B(5, 3) and C(2, 7).
2. Construct the locus of points equidistant from A and B.
3. Construct the locus of points equidistant from AB and AC.
4. Locate the point P such that PA = PB and P is equidistant from AB and AC.
5. Measure and record the length PA in cm. 

Answer 1

 
Steps of Construction:

1. Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA
2. Now we should join points A and B. Draw perpendicular bisector l of AB. Then, l is the locus of points which are equidistant from A and B.
3. Now we should join A and C. Also draw the angle bisector m of ∠CAB. Then, m is the locus of points equidistant from AB and AC.
4. Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P. P is the required point. Since P lies on the perpendicular bisector of AB. Therefore, P is equidistant from A and B.
   Again,
   Since P lies on the angle bisector of angle A.
   Therefore, P is equidistant from AB and AC.
5. On measuring, the length of PA = 2.5 cm  

Answer 2

1. Plot the points A(1, 1), B(5, 3) and C(2, 7) as shown.
2. Join points A and B. Draw right bisector l of AB. Then, l is the locus of points equidistant from A and B.
3. Join A and C. Now draw the bisector m of ∠CAB. Then, m is the locus of points equidistant from AB and AC.
4. The point of intersection P of right bisector of AB and angle bisector of ∠CAB is the point such that PA = PB and P is equidistant from AB and AC.
5. On measuring PA = 2.5 cm.