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

• 4(u hav to take incenter(1)+excem circle(3))
• 14 years agoHelpfull: Yes(13) No(6)
• 1
  because triangle has only one centroid.
• 14 years agoHelpfull: Yes(12) No(10)
• the number of points equidistant from all the 3 lines is nothing but the centroid of the triangle.
  the triangle can have only 1 centroid.
  answer is 1
• 14 years agoHelpfull: Yes(5) No(3)
• 1
• 14 years agoHelpfull: Yes(3) No(4)
• 1
• 14 years agoHelpfull: Yes(2) No(3)
• OPTION C
  ONE INCENTER AND THREE EXCEM CIRCLE
• 8 years agoHelpfull: Yes(0) No(2)