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

• there is a one in circle exits so ans is 1
• 14 years agoHelpfull: Yes(19) No(9)
• 4
• 14 years agoHelpfull: Yes(16) No(13)
• only one point exits its centroid
• 14 years agoHelpfull: Yes(11) No(7)
• pls somebody give logic.
• 14 years agoHelpfull: Yes(5) No(1)
• 3
• 14 years agoHelpfull: Yes(3) No(22)
• plz explain..
• 14 years agoHelpfull: Yes(2) No(2)
• excentres ar equidistant from the lines formmed by extending the 3 sides of the triangle both ways.
  
• 10 years agoHelpfull: Yes(1) No(0)
• ans is 4 because 1 for incentre and other 3 for excentres .So total no of points equidistant from the 3 lines is 4.
  Note:-Every triangle has three excenters and three excircles.The excircle which is tangent to one side of the triangle and the extensions of the other two sides and the centre of the excircle is known as excentre
• 9 years agoHelpfull: Yes(1) No(0)
• plz explain it
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• There are 4 such points. One point is the incenter of the triangle.
  3 are the excenters with respect to each angle of the triangle. In the picture given,
  below I is the Incenter and JA, JB, JC are excenters.
  
  incenter-excenter
• 5 years agoHelpfull: Yes(0) No(0)