Question 1209028
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The side connecting points A = (0,0) and B = (0,4) is vertical y-axis.


The third vertex C lies on the perpendicular to AB in its midpoint.


The midpoint of AB is  the point (0,2).


The perpendicular to AB in its midpoint is horizontal line y= 2.


The altitude of this equilateral triangle has the length  {{{(a*sqrt(3))/2}}} = {{{(4*sqrt(3))/2}}} = {{{2*sqrt(3)}}}.


This altitude is the distance of the third vertex C from y-axis - hence, it gives x-coordinate of vertex C.


Thus, the point C, which is the third vertex of the triangle, has coordinates  (x,y) = (+/-{{{2*sqrt(3)}}},{{{2}}}).    


So, one version is when triangle ABC lies in the first quadrant. Then its third vertex is  (x,y) = ({{{2*sqrt(3)}}},{{{2}}}).


Another version is when triangle ABC lies in the second quadrant. Then its third vertex is  (x,y) = ({{{-2*sqrt(3)}}},{{{2}}}).


So, there are two possible equilateral triangles ABC and two different positions for vertex C.
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Solved.


All you need to know and to use to solve this problem is basic elementary properties of equilateral triangles.


By knowing it and using it, &nbsp;you solve this problem &nbsp;MENTALLY &nbsp;in &nbsp;10 &nbsp;or &nbsp;20 seconds.