Question

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

(i) Plot the points A(-5, -5), B(5, -4), C(4, 8) and D(-4, 6).

(ii) Construct the locus of points equidistant from B and C.

(iii) Construct the locus of points equidistant from AB and AC.

(iv) Locate the point P such that PB = PC and P is equidistant from AB and AC.

Then P lies in

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. fourth quadrant

Answer 1

The correct option is B

second quadrant

Take 1 cm = 1 unit on both axes.

(i) Plot the given points A(-5, -5), B(5, -4), C(4, 8) and D(-4, 6). Join AB, BC, AD and CD.

ii) Construct the perpendicular bisector 'm' of BC.

(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. So, we construct the perpendicular bisector of BC).

(iii) Construct the bisector 'l' of ∠ABC.

(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. So, we construct the bisector for angle B).

(iv) P is the point of intersection of the perpendicular bisector 'm' of the line segment BC and the bisector 'l' of ∠ABC.

From the constructions, we see that P lies in the second quadrant.