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Q3. 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
Read Solution (Total 7)
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- answer is 4......
there will be three circle which can drawn outside the circle taking the sides of the triangle extended as tangents.
hence total points will be 1+3= 4 points.
- 14 years agoHelpfull: Yes(8) No(1)
- 4
one point ie incenter of triangle
other thee point is obtained by xcenter i.e. by angle bisector of exterior angle of three side
hence 1+3=4 - 14 years agoHelpfull: Yes(7) No(1)
- ans is 1 as in a equilateral triangle centroid and orthogonal centroid both are same point
- 14 years agoHelpfull: Yes(4) No(5)
- 1
- 14 years agoHelpfull: Yes(2) No(4)
- 1
- 14 years agoHelpfull: Yes(2) No(4)
- 1
- 14 years agoHelpfull: Yes(2) No(4)
- 1
- 14 years agoHelpfull: Yes(2) No(3)
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