Question

Two vertices of an equilateral triangle are (0,0) and (3 ,root 3) then find the vertices of the third vertice​

Answer 1

SOLUTION:-

O (0,0) & A (3,√3) be the given points and let B (x,y) be the third vertex of equilateral ∆OAB.

⟹ OA = OB = AB

⟹ OA² = OB² = AB²

So,

OA² = (3-0)²+(√3-0)² = 12

OB² = x²+y²

AB² = (x-3)²+(y-√3)² = x²+y²-6x-2√3y+12

_____________________

OA² = OB² = AB²

⟹ OA² = OB² & OB² = AB²

⟹ x²+y² = 12

And also,

x²+y² = x²+y²-6x-2√3y+12

⟹ x²+y² = 12 & 6x+2√3y = 12

⟹ x²+y² = 12 & 3x+√3y = 6

⟹ x²+(6-3x/√3)² = 12

⟹ 3x²+(6-3x)² = 36

⟹ 12x²-36x = 0

⟹ x = 0,3.

_____________________

Therefore,

x = 0 ⇒ √3y = 6 =6/√3

⇒ 2√3 [putting x = 0 in 3x +√3y = 6]

x = 3 ⇒ 9+√3y = 6

⇒ y = 6-9/√3

⇒ -√3 [putting x = 3 in 3x +√3y = 6]

_____________________

Hence, the co-ordinates of the third vertex B are,

➪(0,2√3) or (3,-√3).