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

• The answer is 4
  1 point is I ,the incentre of triangle and the three are Ja, Jb and Jc from the following figure:
  http://starcareer.in/media/triangle.png
  All these points are equidistant from the three lines.
• 13 years agoHelpfull: Yes(53) No(7)
• The correct answer is four. One incentre and three excentres. All are equidistant from the three lines. I am sure about the answer.
  
• 13 years agoHelpfull: Yes(13) No(2)
• one and that is incentre of triangle.
• 13 years agoHelpfull: Yes(6) No(19)
• how can it be 4points... i think it is only one as only incentre or centroid wil be equidistant
• 12 years agoHelpfull: Yes(4) No(5)
• http://files.testfunda.com//Content/ZeusToolsAssets/Content/Questions/Images/FullSize/eada1ab1-d01c-467d-8a43-ab5b46977390.jpg
• 10 years agoHelpfull: Yes(4) No(2)
• only 1 point i.e. incentre is possible.
• 13 years agoHelpfull: Yes(3) No(22)
• i have encountered dis type of questn ... which topic does dis belong to
  
• 12 years agoHelpfull: Yes(1) No(4)
• the answer will be option a) because there are only a equidistant point that lies incentre of the given triange.
  
• 8 years agoHelpfull: Yes(0) No(0)