Question

If two vertices of an equilateral triangle be (0,0) (3,root3) find the third vertex

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

The third vertex of the equilateral triangle is, $(0,2 \sqrt{3}) \ or (3,-\sqrt{3})$

What is the vertex of a triangle?

• A point where any two sides of a triangle meet, is called a vertex of a triangle.

How do you find the third vertex of an equilateral triangle?

• Substituting the value of y in equation (1) and equating it to 26. Solving the quadratic equation using the quadratic formula, [-b ± √(b2 - 4ac)]/2a.
• Hence, the coordinates of the third vertex of the equilateral triangle are ([1 ± √(3)] / 2, [7 ± 5√(3)] / 2).

According to the question:

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).

$\sqrt{\left(x^{2}+y^{2}\right)}=\sqrt{\left(3^{2}+3\right)}=\sqrt{(x-3)^{2}+(y-\sqrt{3})^{2}}\\ x^{2}+y^{2}=12\\ x^{2}+9-6 x+y^{2}+3-2 \sqrt{3 y}=12\\ 24-6 x-2 \sqrt{3 y}=12 \\ -6 x-2 \sqrt{3 y}=-12 \\ 3 x+\sqrt{3 y}=6 \\ x=\frac{6-\sqrt{3 y}}{3}$

$\frac{(6-\sqrt{3 y})}{3} 2+y^{2}=12\\ \frac{(36+3 y 2-12 \sqrt{3 y})}{9}+y^{2}=12\\ 36+3 y^{2}-12 \sqrt{3 y}+9 y 2=108 \\ -12 \sqrt{3 y}+12 y^{2}-72=0 \\ \sqrt{-3 y}+y 2-6=0 \\ (y-2 \sqrt{3})(y+\sqrt{3})=0 \\ y=2 \sqrt{3} or \sqrt{-3}$

If $y=2 \sqrt{3}, x=(6-6) / 3=0$

If $y=2 \sqrt{3}, x=(6-6) / 3=0$

So,the third vertex of the equilateral triangle= $(0,2 \sqrt{3}) \ or (3,-\sqrt{3})$

Hence, the third vertex are  $(0,2 \sqrt{3}) \ or (3,-\sqrt{3})$

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