Question

show that the three points ( 1,1) , ( root3, root -3) and ( -1, -1)are the vertices of an equilateral triangle.

Answer 1

using distance formula

Answer 2

Answer:

AB=BC=AC

Step-by-step explanation:

$A= (1,1)$

$B=(\sqrt{3},-\sqrt{3})$

$C=(-1,-1)$

To find AB use distance formula :

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

$(x_1,y_1)=(1,1)$

$(x_2,y_2)=(\sqrt{3},-\sqrt{3})$

Substitute the values in the formula :

$AB=\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2}$

$AB=2.828$

To Find BC

$(x_1,y_1)=(\sqrt{3},-\sqrt{3})$

$(x_2,y_2)=(-1,-1)$

Substitute the values in the formula :

$BC=\sqrt{(-1-\sqrt{3})^2+(-1+\sqrt{3})^2}$

$BC=2.828$

To Find AC

$(x_1,y_1)= (1,1)$

$(x_2,y_2)= (-1,-1)$

Substitute the values in the formula :

$AC=\sqrt{(-1-1)^2+(-1-1)^2}$

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

$AC=\sqrt{4+4}$

$AC=\sqrt{8}$

$AC=2.828$

AB=BC=AC

Since all the sides of the triangle are equal .

So, the given three points are the vertices of equilateral triangle.