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Show that (a, a), (- a, - a) and  \left ( -\sqrt{3}a,\sqrt{3}a \right )  are vertices of an equilateral triangle.

 

 

 

 
 
 
 
 

Answers (1)

Given 

A=(a,a), B=(-a,-a), C=(-\sqrt{3}a,\sqrt{3}a)

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

         =\sqrt{4a^2 + 4a^2}

         =\sqrt{8a^2 }

AB= 2\sqrt{2}a

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

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

         =a\sqrt{1+3-2\sqrt{3} + 1+3+2\sqrt{3}}

         =a\sqrt{8}

BC=2\sqrt{2}a

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

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

          =a\sqrt{3+1+2\sqrt{3} + 3+1-2\sqrt{3}}

          =a\sqrt{8}

CA=2\sqrt{2}a

\therefore AB=BC=CA

It forms an equilateral triangle.

 

Posted by

Ravindra Pindel

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