Mathematics
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°.
Hence,
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Locus
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Answer
The figure is shown below:
(i) We know that the locus of points equidistant from two lines is equal to the perpendicular bisector of angle between the lines.
From figure,
The locus of points equidistant from BA and BC is angle bisector of ABC i.e. BX.
(ii) We know that the locus of points equidistant from two points is the perpendicular bisector of the line joining the two points
From figure,
The locus of points equidistant from B and C is the perpendicular bisector of BC i.e. YZ.
(iii) From figure,
YZ and BX meet at point P.
Hence, P is the point which satisfies above two loci and PC = 5.1 cm.
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