I have an equilateral triangle with known coordinations of one vertex (a,b). I know that each side's length is t and the coordinations of the other vertexes (x1, y1), (x2, y2). There is another point (c, d) which creates with (a,b) a segment in length k that is perpendiculars to the opposite side of (a,b). I need to express x1, y1, x2 and y2 with the given parameters a, b, c, d, t and k. I started with four equations: Because we have perpendicular segments, we can say that their slopes's product equals -1.
1)$$\frac{d-b}{a-b}= \frac{x2-x1}{y2-y1}$$ It is also known that the distance between each two vertexes is t, so expressing each side with the given coordinations gives us another three equations:
2) $$(a-x1)^2 + (b-y1)^2 = t^2$$ 3) $$(a-x2)^2 + (b-y2)^2 = t^2$$ 4) $$(x2-x1)^2 + (y2-y1)^2 = t^2$$
I tried solving the equation and got the following answers:
$$x1 = a-\frac{t(d-b+\sqrt{k^2-3(d-b)^2})}{2k}$$ $$y1 = b+\frac{t((d-b)\sqrt{k^2-3(d-b)^2}-(a-c)^2)}{2k(a-c)}$$ $$x2 = a+\frac{t(d-b-\sqrt{k^2-3(d-b)^2})}{2k}$$ $$y2 = b+\frac{t(2(d-b)(a-c)-(a-c)^2+(d-b)\sqrt{k^2-3(d-b)^2})}{2k(a-c)}$$
I tried to substitute some numbers for the parameters but the 3 vertexesdposent create an equilateral triangle.
I'll be really glad if you could help me. edit: still havnt figuered it out anyhelp will be welcome.
Say that we are given $A(a,b)$. Then as $AB=AC=BC=t$, $$C(a+t, b)$$ and $$B\left(a+\frac t2, b+\frac{t\sqrt3}4\right)$$ by Pythagoras, as the $x$-coordinate of $B$ is halfway between that of $A$ and $C$.
Can you do something similar if $B$ or $C$ was the given vertex?