A and B are fixed points such that AB=2a. The vertex C of `DeltaABC` such that `cotA+cotB`=constant. Then locus of C is
A
straight line perpendicular to AB
B
straight line parallel to AB
C
circle
D
none of these
Text Solution
Verified by Experts
The correct Answer is:
B
`cot A +cotB=` constant `rArr (AD)/(CD)+(BD)/(CD)=` constant, (where D is the foot of perpendicular from C to AB) `rArr (AD+BD)/(CD) =` constant `rArr (2a)/(CD) =` constant `rArr CD =` constant C is moving at a constant distance from AB `:.` locus of C is a line parallel to AB
A and B are fixed points such that AB=2a .The vertex C of Delta ABC such that cot A+cot B= constant.Then locus of C is
A and B are fixed points of PA+PB=K (constant) and K>AB then the locus of P is
A and B are two fixed points. The locus of P such that in Delta PAB,(sin B)/(sin A) is a constant is (!=1)
Let A(2, 0) and B(-2, 0) are two fixed vertices of DeltaABC . If the vertex C moves in the first quadrant in such a way that cotA+cotB=2, then the locus of the point C is
The base AB of a triangle is fixed and its vertex C moves such that sin A =k sin B (kne1) . Show that the locus of C is a circle whose centre lies on the line AB and whose radius is equal to (ak)/((1-k^2)) , a being the length of the base AB.
If A,B,C are the angles off DeltaABC , then cotA*cotB+cotB*cotC+cotC*cotA =
In Delta ABC, vertex B and C are fixed and vertex A is variable such that 2sin((B-C)/(2))=cos((A)/(2)) then locus of vertex A is
If A+B+C=pi/2 , show that : cotA+cotB+cotC=cotA cotB cotC
CENGAGE-COORDINATE SYSTEM-Multiple Correct Answers Type