Question

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

Answer 1

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Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

P(a,a)

Q(-a , -a)

R(-√3a , √3a)

$\large{\sf\:{PQ=\sqrt{(-a-a)^2+(-a-a)^2}}}$

$\large{\sf\:{PQ=\sqrt{(4a)^2+(4a)^2}}}$

$\large{\sf\:{PQ=2\sqrt{2}a}}$

Now

$\large{\sf\:{QR=\sqrt{(-\sqrt{3}a+a)^2+(\sqrt{3}a+a)^2}}}$

$\large{\sf\:{QR=\sqrt{3a^2+a^2-2\sqrt{3}a^2+3a^2+a^2+2\sqrt{3}a^2}}}$

$\large{\sf\:{QR=\sqrt{8a^2}}}$

$\large{\sf\:{QR=2\sqrt{2}a}}$

Now

$\large{\sf\:{PR=\sqrt{(-\sqrt{3}a-a)^2+(\sqrt{3}a-a)^2}}}$

$\large{\sf\:{PR=\sqrt{3a^2+a^2+2\sqrt{3}a^2+3a^2+a^2-2\sqrt{3}a^2}}}$

$\large{\sf\:{PR=\sqrt{8a^2}}}$

$\large{\sf\:{PR=2\sqrt{2a}}}$

Therefore,

PQ = QR = PS = 2√2a

Therefore,

∆PQR is an equilateral Triangle