Question

show that the three points (a,a) , (-a,-a) and (-a√3 , a√3) are vertices of an equilateral triangle

Answer 1

A = (a,a)

B = (-a,-a)

C = (-√3a,√3a)

AB=√(-a-a)²+(-a-a)²

       √4a²+4a²

         √8a²

         2√2 a

BC=√(-√3a+a)²+(√3a+a)

        √(1-√3)²a²+(√3+1)²a²

        a√[ 1+3-2√3+3+1+2√3]

        a[ √8]

        2√2a

CA = √(-√3a-a)²+(√3a-a)²

         √(-√3-1)²a²+(√3-1)²a²

         a√[3+1+2√3+3+1-2√3]

         a√8

         2√2a

∴AB = BC = CA

  it forms a eqilateral traingle

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Answer 2

Distance between all vertices are same. So, the three points are vertices of an equilateral triangle.

Step-by-step explanation:

Consider the three points are A(a,a) , B(-a,-a) and C(-a√3 , a√3).

We need to show that the given points are vertices of an equilateral triangle.

Equilateral triangle: All sides of an equilateral triangle are congruent.

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using  distance formula we get

$AB=\sqrt{(-a-a)^2+(-a-a)^2}=\sqrt{4a^2+4a^2}=\sqrt{8a^2}=2\sqrt{2}a$

$BC=\sqrt{(-a\sqrt{3}+a)^2+(a\sqrt{3}+a)^2}$

$BC=\sqrt{(3a^2-2\sqrt3 a^2 +a^2)+(3a^2+2\sqrt3 a^2 +a^2)}=\sqrt{8a^2}=2\sqrt{2}a$

$AC=\sqrt{(-a\sqrt{3}-a)^2+(a\sqrt{3}-a)^2}$

$AC=\sqrt{(3a^2+2\sqrt3 a^2 +a^2)+(3a^2-2\sqrt3 a^2 +a^2)}=\sqrt{8a^2}=2\sqrt{2}a$

Since,

$AB=BC=AC=2\sqrt{2}$

Therefore, the three points are vertices of an equilateral triangle.

#Learn more

The distance between the point (4,3) and the origin is ​

https://brainly.in/question/13992695