Question

If $A(3,2),$ $B(-3,2),C(0,h)$ are the vertices of an equilateral triangle and $h<0$, then the value of $h^2+12\sqrt{3}$ is

Answer 1

$A(3,2)$, $B(-3,2)$, $C(0,h)$ are the vertices of an equilateral triangle.
$\Rightarrow AB=BC=AC$
$\Rightarrow A{B}^2=B{C}^2=A{C}^2$
$\Rightarrow 36=9+(h-2{)}^2=9+(h-2{)}^2$
$\Rightarrow (h-2{)}^2=27$
$\Rightarrow h-2=\pm \sqrt{27}$
$\Rightarrow h=2\pm 3\sqrt{3}$
Since $h<0$
$\Rightarrow h=2-3\sqrt{3}$
$\Rightarrow h^2=4+27-12\sqrt{3}$
$\Rightarrow h^2+12\sqrt{3}=31$