Latest TCS Aptitude Question SOLUTION: Given 3 lines in the plane such that the point of intersection form a triangle with sides of lenght 20,20 and 18,the number of points equidistance from all the 3 lines is a.)1 b.)

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40010596) TCS question Keep an EYE

Given 3 lines in the plane such that the point of intersection form a triangle with sides of lenght 20,20 and 18,the number of points equidistance from all the 3 lines is

a.)1
b.)4
c.)0
d.)3

Asked In TCS Pankaj Jain (11 years ago)

Solved PARAG KAMRA Read Solution (4)

Is this Puzzle helpful?   (19)   (5) Submit Your Solution

Read Solution (4) :

PARAG KAMRA  (11 years ago)

Point to remember is that incenter and excenter are the points which are equidistant from the three sides of a triangle. Incenter always lie inside the triangle as it is the center of incircle and the excenters always lie outside the circle as they are center of three excircles.
b)4 (ans)

Is this solution Helpfull? Yes (43) | No (3)  

ragini devi  (11 years ago)

in a triangle sum of any 2 sides should be greater than the 3rd side. so 20,20,18 do not constitute a triangle

Is this solution Helpfull? Yes (2) | No (23)  

Sourabh Malu  (11 years ago)

i think ans is C) becoz all lines are making a triangle

Is this solution Helpfull? Yes | No (16)  

shivi misra  (11 years ago)

I THINK THE ANS TO BE c BEC INTERSECTION OF LINES FORM A TRIANGLE THATS WHY NO POINT IS EQUIDISTANCE FOR ALL 3

Is this solution Helpfull? Yes | No (9)  

Please Login to Read Solution.