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An equilateral triangle has one vertex at $(0,0)$ and another at $(3, \sqrt{3})$. What are the coordinates of the third vertex?
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1805 Upvotes
Verified Answer
The correct answer is:
$(0,2 \sqrt{3})$ or $(3,-\sqrt{3})$
Let $A B C$ is equilateral triangle with $A(0,0)$ and $B(3, \sqrt{3})$ and $C$ to be known.
$A B=\sqrt{(3-0)^{2}+(\sqrt{3}-0)^{2}}=\sqrt{9+3}=\sqrt{12}$
Take option (a) i.e. $C(0,2 \sqrt{3})$
$C A=\sqrt{0^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
Take option (b) i.e. $C(3,-\sqrt{3})$
$C A=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(0)^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
Both option (a) and (b) are correct.
$A B=\sqrt{(3-0)^{2}+(\sqrt{3}-0)^{2}}=\sqrt{9+3}=\sqrt{12}$
Take option (a) i.e. $C(0,2 \sqrt{3})$
$C A=\sqrt{0^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
Take option (b) i.e. $C(3,-\sqrt{3})$
$C A=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(0)^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
Both option (a) and (b) are correct.
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