Question

If 2 vertices of equilateral triangle be (0,0) and (3,√3).Find third vertex.

Answer 1

let B(x,y) be the 3rd vertex of triangle OAB. then

OA=OB=OC

OA^2=OB^2=AB^2

OA^2=(3-0)^2+(√3-0)^2=12

OB^2=x^2+y^2

AB^2=(X-3)^2+(Y-√3)^2

AB^2=X^2+Y^2-6X-2√3Y+22

THEREFORE,OA^2=0B^2=AB^2

OA^2=OB^2=AB^2

X^2+Y^2=12

X^2+Y^2=X^2+Y^2-6X-2√3Y+12

X^2+Y^2=12 AND 6X+2√3Y=12

X^2+Y^2=12 AND 3X+√3Y=6

X^2+{6-3X/√3}^2=12.therefore (3X+√3Y=6.SO,Y=6-3X/√3)

3X^2+(6-3X)^2=36

12X^2-36X=0

X=0,3

X=0==. √3Y=6. == Y=6/√3=2√3
X=3==. 9+√3Y=6==Y=6-9/√3= -√3

THEREFORE THIRD VERTEX IS (0,2√3) OR (3,-√3).
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