Question

Show that the point (-3,-3),(3,3)and( -3√3,3√3) are the vertices of an equilateral triangle

Answer 1

$\textbf{Given:}$

$\text{Points are (-3,-3),(3,3) and (-3\sqrt{3},3\sqrt{3}) }$

$\textbf{To prove:}$

$\text{They form an equilateral triangle}$

$\textbf{Solution:}$

$\text{Let the given points be}$

$A(-3,-3),\;B(3,3),\,\text{and}\,C(-3\sqrt{3},3\sqrt{3})$

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

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

$AB=\sqrt{(-6)^2+(-6)^2}$

$AB=\sqrt{36+36}$

$AB=\sqrt{72}$

$\implies\boxed{AB=6\sqrt{2}}$

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

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

$BC=\sqrt{9+27+18\sqrt{3}+9+27-18\sqrt{3}}$

$BC=\sqrt{36+36}$

$BC=\sqrt{72}$

$\implies\boxed{BC=6\sqrt{2}}$

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

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

$AC=\sqrt{9+27-18\sqrt{3}+9+27+18\sqrt{3}}$

$AC=\sqrt{36+36}$

$AC=\sqrt{72}$

$\implies\boxed{AC=6\sqrt{2}}$

$\implies\,AB=BC=AC$

$\therefore\textbf{The given points form an equilateral triangle}$

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Answer 2

Step-by-step explanation:

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