Question

If the points $(1,1),(-1,-1)$ and $(-\sqrt{3},k)$ are vertices of an equilateral triangle find the value of $k$.

Answer 1

$A(1,1),B(-1,-1),C(-\sqrt{3},k)$

$AB=AC$

$\Rightarrow \sqrt{4+4}=\sqrt{(1+\sqrt{3}{)}^2+(1-k{)}^2}$

$\Rightarrow 8=1+3+2\sqrt{3}+(1-k{)}^2$

$\Rightarrow 4-2\sqrt{3}=(1-k{)}^2$

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

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

$\therefore 1-k=\sqrt{3}-1,$ or $1-k=1-\sqrt{3}$

$\Rightarrow k=2-\sqrt{3}$ or $-k=\sqrt{3}$

$\therefore k=2-\sqrt{3},k=\sqrt{3}$