Find $B=(x,y)$ so triangle is equilateral Let $O=(0,0)$, $A=(3,4)$ and $B=(x,y)$ be three points in $xOy$. Find real numbers $x$ and $y$, so that $OAB$ is an equilateral triangle.
I'm really struggling with this one, can someone help?
 A: Hint:
Look at the figure, where the distance $OA$ is known.

What does it suggest ?
A: We need to have $|\overline{OB}|=|\overline{AB}|=|\overline{OA}|=5$. Thus from
$$ |\overline{OB}|^2=x^2+y^2=25$$
and $$|\overline{AB}|^2=(x-3)^2+(y-4)^2=25$$we reach
$$x^2-6x+9+y^2-12y+16=x^2+y^2 $$
or $$6x+12y=25\text{ or }x=2y-25/2.$$
Replace this value to $x^2+y^2=25$:
$$4y^2+50y+625/4+y^2=25$$
or $$y^2+10y-\frac{150}{4}=0 .$$
The solutions of this equation are
$$y=\frac{-1\pm5\sqrt{10}}{2} $$
The rest is obvious.
A: Hint: Use the distance formula $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
You know that the distance between $O$ and $A$ is $5$. Now solve for this system of equations:
$$5=\sqrt{(x-3)^2+(y-4)^2}$$
$$5=\sqrt{x^2+y^2}$$
A: Alternatively, the distance $AO$ is $5$ and the point $C$ must lie on the line that cuts $AB$ perpendicularly at its center, because $AC=BC$. Also, the point $C(x,y)$ must lie at a distance of $5$ from the origin:
$$\begin{cases}x^2+y^2=5^2 \\ y=-\frac43x+\frac{25}{8}\end{cases} \Rightarrow \begin{cases} 4x^2-12x-39=0 \\ y=-\frac43x+\frac{25}{8}\end{cases} \Rightarrow \begin{cases} x=\frac32\pm 2\sqrt{3} \\ y=2\mp \frac32 \sqrt{3}\end{cases}.$$
A: Correct me if wrong.
Length $OA = \sqrt{3^2+4^2} =5,$ the side length of the equilateral triangle.
Height on $OA$; $h = \sqrt{5^2-(5/2)^2}= \sqrt{75/4}=(5/2)√3.$
Slope of $OA$:  $m:=4/3.$
Slope of a line perpendicular to $OA$: $m'=-3/4$.
Midpoint of OA: (3/2,2).
Direction vector normalized with slope $-3/4$: 
$\vec d= (1/5)(-4,3)$.
$\vec r = (3/2,2) + t (1/5)(-4,3).$
$t=^+_- h$ will give the $2$ vertices $B.$
