Given 3 lines in the plane such that the points of intersection from a triangle with sides of length 20, 20 and 31, the number of points equidistant from all the 3 lines is

• 4
• 3
• 0
• 1

• the ans will be 4..there are four points that are equidistant from all 3 lines..1 incentre and 3 excentre...thank you..hve a nice day
• 13 years agoHelpfull: Yes(15) No(4)
• 4 points
  one will be incentre and 3 excentres.
• 13 years agoHelpfull: Yes(5) No(2)
• it ans will be 1 .....
• 13 years agoHelpfull: Yes(2) No(5)
• the ans isssssss 1
• 13 years agoHelpfull: Yes(2) No(5)
• it will be incenter so ans is 1
  
• 13 years agoHelpfull: Yes(1) No(3)
• which one is correct 4 or 1???
• 13 years agoHelpfull: Yes(1) No(0)
• since it is equidistant....the required point is the orthocenter
  so ans is 1
• 13 years agoHelpfull: Yes(0) No(1)