Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 19

only one i.e., Circumcenter
9 years agoHelpfull: Yes(9) No(0)
triangle in the plane so only 1 point (centre)
9 years agoHelpfull: Yes(4) No(2)
only 1 point
9 years agoHelpfull: Yes(4) No(1)
when the three lines are drawn to for a triangle the poins euiditant are 1 incentre and 3 excentres so total no of points equidistant are 3+1=4 ans:4
9 years agoHelpfull: Yes(1) No(2)
excentres are not equidistant from 3 sides of triangle. Only incentre is equidistant from 3 sides of triangle. So the ans is one
9 years agoHelpfull: Yes(1) No(0)
No of points equidistant from lines=4
No of point equidistant from segments=1
9 years agoHelpfull: Yes(1) No(0)
Making pyramid, we can get two points one upside and one downside the plane of triangle.
9 years agoHelpfull: Yes(0) No(4)
there are 4 such points.One point will be the incentre of the circle and the 3 of them will be the excentre.
9 years agoHelpfull: Yes(0) No(2)

Given 3 lines in the plane such that the points of intersection form a triangle with sides of length 19, 19 and 19, the number of points equidistant from all the 3 lines is