Given 3 lines in the plane such that the points on intersection form a triangle with sides of length 20, 20 and 22 the number of points equidistant from all the 3 lines is
4
1
0
3

• there would be 4 points
• 12 years agoHelpfull: Yes(6) No(7)
• ans is 4.bcz...for any type(in lengths) of triangle the equidistance points from their sides are 3 ex-centers & 1-incenter
• 12 years agoHelpfull: Yes(3) No(2)
• 0
  if this was an equilateral triangle then the answer would have been 1(the in-center).
• 12 years agoHelpfull: Yes(2) No(1)
• ans is 1 and the point is the incenter of the triangle
  
• 12 years agoHelpfull: Yes(1) No(3)