Question

if points (0,0),(3,√3),(x, y) form an equilateral triangle, then (x,y)=

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Answer 1

Answer :

The third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Step-by-step explanation:

Correct Question :

If two vertices of an equilateral triangle be (0,0) (3,√3), Find the third vertex.

Given :

Two vertices of an equilateral triangle are (0, 0) and (3, √3).

To Find :

The third vertex.

Solution :

★ Consider the -

the third vertex of the equilaterlateral triangle as - (x, y)

★ Remember :

Distance between (0, 0) and (x, y) = Distance between (0, 0) & (3, √3) = Distance between (x, y) and (3, √3).

So,

Values in Equation,

$\bigstar\;\boxed{\sf \sqrt{(x2 + y2)} = \sqrt{(32 + 3)}= \sqrt{[(x - 3)2} + (y - \sqrt3)2]}}$

Now,

$\sf\\\implies x2 + y2 = 12\\ \\\sf \implies x2 + 9 - 6x + y2 + 3 - 2\sqrt3y = 12\\ \\\sf \implies 24 - 6x - 2\sqrt3y = 12\\ \\\sf \implies - 6x - 2\sqrt3y = - 12\\ \\\sf \implies 3x + \sqrt3y = 6\\ \\\sf \implies x= \dfrac{(6 -\sqrt3y)}{3}$

So,

$\sf\\\implies \dfrac{(6 - \sqrt3y)}{3}2 + y^2 = 12\\ \\\sf \implies \dfrac{(36 + 3y^2 - 12 \sqrt3y)}{9} + y^2= 12\\ \\\sf \implies 36 + 3y2 - 12\sqrt3y + 9y^2 = 108\\ \\\sf \implies - 12\sqrt 3y + 12y^2 - 72 = 0\\ \\\sf \implies \sqrt-3\;\sqrt3y + y^2 - 6 = 0\\ \\\sf \implies (y - 2\sqrt3)(y + \sqrt3) = 0\\ \\\sf \implies y = 2\sqrt3\;and\;- \sqrt3$

So,

$\sf y = 2\sqrt3,\;x = \dfrac{(6 - 6)}{3} = 0\\ \\\sf y = -\sqrt3,\;x = \dfrac{(6 + 3)}{3} = 3$

∴ The third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Answer 2

Answer:

here

Step-by-step explanation:

ANSWER

Two vertices of an equilateral triangle are (0,0)and(3,√3).

Let the third vertex of the equilateral 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+3−23y=12

24−6x−23y=12

−6x−23y=−12

3x+3y=6

x=36−3y

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