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प्रश्न
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
- Plot the points A(1, 1), B(5, 3) and C(2, 7).
- Construct the locus of points equidistant from A and B.
- 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.
- Measure and record the length PA in cm.
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उत्तर १
Steps of Construction:
- Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA
- 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.
- 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.
- 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. - On measuring, the length of PA = 2.5 cm
उत्तर २
- Plot the points A(1, 1), B(5, 3) and C(2, 7) as shown.
- Join points A and B. Draw right bisector l of AB. Then, l is the locus of points equidistant from A and B.
- Join A and C. Now draw the bisector m of ∠CAB. Then, m is the locus of points equidistant from AB and AC.
- 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.
- On measuring PA = 2.5 cm.
संबंधित प्रश्न
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
- equidistant from BA and BC.
- 4 cm from M.
- 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.
Construct a triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.
Using ruler and compasses construct:
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of point equidistant from A and C.
(iii) a circle touching AB at A and passing through C.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Without using set squares or protractor.
(i) Construct a ΔABC, given BC = 4 cm, angle B = 75° and CA = 6 cm.
(ii) Find the point P such that PB = PC and P is equidistant from the side BC and BA. Measure AP.
Ruler and compass only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
Given ∠BAC (Fig), determine the locus of a point which lies in the interior of ∠BAC and equidistant from two lines AB and AC.
