Question

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

(i) Plot the points A(-2, 4), B(2, 6) and C(3, -6).

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

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

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

Then the length(approximately) of PA in cm is

• A. 5.2 cm
• B. 5.8 cm
• C. 6.2 cm
• D. 8.1 cm

Answer 1

The correct option is B

5.8 cm

Take 1 cm = 1 unit on both axes.

(i) Plot the given points A(-2, 4), B(2, 6) and C(3, -6). Join AB, BC and AC to form triangle ABC.

(ii) Construct the perpendicular bisector of AC.

(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 AC).

(iii) Construct the bisector of ∠BCA.

(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 C).

(iv) P is the point of intersection of the perpendicular bisector of the line segment AC and the bisector of ∠BCA

On measuring, we get, length of PA = 5.8 cm approx.