Mathematics
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Locus
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Answer
Steps of construction :
Draw a line segment AB = 6 cm
From A and B as centers and radius 9 cm, make two arcs which intersect each other at C.
Join CA and CB.
Draw the perpendicular bisector of BC.
With A as center and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
Hence, P is the required point which is equidistant from B and C and at a distance of 4 cm from A.
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The locus of point which is equidistant from two non-parallel lines AB and CD is :
perpendicular to AB.
perpendicular to CD.
bisector of angle between AB and CD.
perpendicular bisector of CD.
Three isosceles triangles PBC, QBC and RBC are on the same base, then :
P, Q and R are collinear.
△PQR is isosceles triangle.
Q lies on the circumference of a circle with BC as diameter.
Q is mid-point of line segment PR.