Question

If two vertices of an equilateral triangle are (0,0) and (3,√3) find the thord vertice

Answer 1

Two vertices of an equilateral triangle are (0, 0) and (3, √3). Let the third vertex of the equilaterla triangle be (x, y) Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3) √(x2 + y2) = √(32 + 3) = √[(x - 3)2 + (y - √3)2] x2 + y2 = 12 x2 + 9 - 6x + y2 + 3 - 2√3y = 12 24 -  6x - 2√3y = 12 - 6x - 2√3y = - 12 3x + √3y = 6 x = (6 - √3y) / 3 ⇒ [(6 - √3y)/3]2 + y2 = 12 ⇒ (36 + 3y2 - 12√3y) / 9 + y2 = 12 ⇒ 36 + 3y2 - 12√3y + 9y2 = 108 ⇒ - 12√3y + 12y2 - 72 = 0 ⇒ -√3y + y2 - 6 = 0 ⇒ (y - 2√3)(y + √3) = 0 ⇒ y = 2√3 or - √3 If y = 2√3, x = (6 - 6) / 3 = 0 If y = -√3, x = (6 + 3) / 3 = 3 So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Answer 2

Answer:

Step-by-step explanation:

The distance between two points is given by √[(x2 - x1)2 + (y2 - y1)2] Let the two vertices of equilateral triangle be A(0, 0) and B(3,√3) Let the third vertex be C(a, b) AB = √[(3 - 0)2 + (√3 - 0)2] = √12 Similarly, AC = √[a2 + b2] BC = √[a2 - 6a + b2 - 2√3b + 12] Hence √[a2 - 6a + b2 - 2√3b + 12] = √12 [Since AB = BC] a2 - 6a + b2 - 2√3b + 12 = 12 12 - 6a - 2√3b = 0 6a = 2√3b + 12 a = (√3b/3) + 2 Also a2 + b2 = 12 Put value of a in above equation and solve to get the value of b.

Your answer will be (0,2 root 3) and (3,- root 3).

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