Question

find the third vertex of the equilateral triangle A(0,0),B(3,√3)

Answer 1

Let 3rd vertex be C(x,y). We have AB = BC = AC
AB =
$\sqrt{ {(3 - 0)}^{2} + {( \sqrt{3} - 0) }^{2} } = \sqrt{12}$
BC =
$\sqrt{ {(x - 3)}^{2} + {(y - \sqrt{3}) }^{2} }$
AC =
$\sqrt{ {x}^{2} + {y}^{2} }$
AB = AC gives
${x}^{2} + {y}^{2} = 12 - - - (1)$
BC = AC gives
${(x - 3)}^{2} + {(y - \sqrt{3)} }^{2} = {x}^{2} + {y}^{2} \\ 3x + \sqrt{3} y = 6 - - - (2)$
Substituting (2) in (1),
${( \frac{6 - \sqrt{3}y }{3}) }^{2} + {y}^{2} = 12 \\ {y}^{2} - \sqrt{3} y - 6 = 0 \\ y = 2 \sqrt{3} \: \: \: or \: \: \: - \sqrt{3}$
Substituting values of y in (2),
$x = 0 \: \: \: or \: \: \: 3$
Thus the 3rd vertex is (0,2√3) or (3,-√3)