Question

two vertex of equilateral triangle are (0,0) and (3,√3) find the third vertex

Answer 1

Let OAB be the equilateral triangle such that O(0,0) and A(3,3‾√33). 

If you sketch it on xy-plane, you would notice that the line segment OA lies on the first quadrant. Another important observation is to notice that the third vertex could lie either on the same vertical line passing through A in the fourth quadrant or exactly on the y-axis i.e. above O. 

Reason that can be associated with the above observation is based on the gradient(slope) calculation of OA. Gradient or slope of OA, m = tan(θ) =3√−03−0 =13√tanθ 3030 13. It clearly tells us that θ =300θ 300, that means the line OA is inclined at an angle of 300300
 in the first quadrant. Each angle in an equilateral triangle must be 600600.  For ∠O∠O to be 600600, the another 300300 must move the side OB into the fourth quadrant and eventually if you draw a line perpendicular to y-axis passing through A down to the fourth quadrant, you would see that the third vertex will lie on this line and in addition, it will be exactly same distance away from x-axis as the vertex A i.e. 3‾√3. Also, it can be proved using alternate interior angle properties that ∠OBA=600∠OBA600. It is pretty much obvious that x-axis is actually a median or an altitude of the equilateral triangle OAB. So, the coordinate of B in the fourth quadrant will be (3,−3‾√33).

Also, we must be careful not to forget that the third side OB can be moved in anti-clockwise direction too. Since OA makes an angle of 300300 with positive x-axis, it makes an angle of 600600 with the y-axis. Clearly, y-axis provides us another base line for the third vertex B. Now, the position of A is exactly the position of median or altitude through vertex A. And we know that median passes through the half way of third side. Median or altitude will meet the y-axis at M(0,3‾√03). It is implied that OB = 2OM, so the third vertex B is given by (0,23‾√023).