TCS Company

Given 3 lines in the plane such that the points of
intersection form a triangle with sides of length 20, 20 and
30, the number of points equidistant from all the 3 lines is

Read Solution (Total 5)

The answer would be 4 - 1 would be incentre and 3 excentres.
I request you to first search this site for the question before your question.

Thanks
12 years agoHelpfull: Yes(5) No(0)

there is only 1 point which is equidistant fron all the three points and that is the incenter of the triangle.
12 years agoHelpfull: Yes(2) No(0)
answer is 4.. 1 incentre and 3 excentre..
12 years agoHelpfull: Yes(2) No(0)
the answer is 1
the incentre of the triangle.
12 years agoHelpfull: Yes(0) No(1)
shivam plz explain it why u take 1 incentre and 3excentre
12 years agoHelpfull: Yes(0) No(0)

TCS Other Question

In a sequence of integers , A(n)= A(n-1)-A(n-2), where A(n) is the nth term in the sequence, n is the integer and n>=3 , A(1)=1, A(2)=1,. calculate S(1000), where S(1000) is the sum of the first 1000 terms what is the reminder when 6^17+7^16 is divided by 7