I have two triangles that are not similar. The Only thing that I know is that AB and C points from triangle 1 are related to $A^1 B^1$ and $C^1$ points from triangle 2.
Based on these inputs I want to find $K^1$ point related to Triangle 2 based on K point related to triangle 1.
In my problem, these triangles have nothing in common. But by rotating, scaling and moving them I reach the level where:
both of these triangles have the same border AB ( $AB = A^1B^1$ ) (border d from the image)
A point is located at the origin of XY coordinate system ( $A = A^1 = [0, 0]$ )
but coordinates of C and $C^1$ are different.
It looks like this:
This image tells it better. I need to find coordinates for K1.
I tried following equalities based on a feeling of a ratio between C, K and C1, K1 points:
for X Coordinate: $\frac{C_x}{K_x} = \frac{C_x^1}{K_x^1}$
same for Y Coordinate $\frac{C_y}{K_y} = \frac{C_y^1}{K_y^1}$
From here i could easily find out X and Y coordinates for K1:
$ K_x^1 = \frac{K_x * C_x^1}{C_x} $
$ K_y^1 = \frac{K_y * C_y^1}{C_y} $
I tested this for multiple points programmatically and it seems correct, but i couldn't find mathematical proof for this equality.
I wonder if there is any method for it already? Unfortunately, I couldn't find anything related.
please correct me if I'm using the wrong terminology
Let's assume points $K$ and $K_1$ have the same barycentric coordinates.
Since beyond OP, this may be of use to others encountering a similar situation, let us look at how to define barycentric coordinates with respect to an arbitrary triangle.
Let's say the vertices of the triangle are $$\vec{p}_1 = \left [ \begin{matrix} x_1 \\ y_1 \end{matrix} \right ], \quad \vec{p}_2 = \left [ \begin{matrix} x_2 \\ y_2 \end{matrix} \right ], \quad \vec{p}_3 = \left [ \begin{matrix} x_3 \\ y_3 \end{matrix} \right ] \tag{1}\label{NA1}$$ In barycentric coordinates, the coordinate axes are $$\hat{u} = \vec{p}_2 - \vec{p}_1, \quad \hat{v} = \vec{p}_3 - \vec{p}_1 \tag{2}\label{NA2}$$ so that point $\underline{p} = (u, v)$ in barycentric coordinates corresponds to point $\vec{p} = (x, y)$ in Cartesian coordinates: $$\vec{p} = \vec{p}_1 + u \hat{u} + v \hat{v} \quad \iff \quad \left\lbrace\begin{aligned} x &= (1 - u - v) x_1 + u x_2 + v x_3 \\ y &= (1 - u - v) y_1 + u y_2 + v y_3 \end{aligned} \right. \tag{3}\label{NA3}$$ Barycentric coordinates $(u, v)$ are within the triangle, if and only if $0 \le u \le 1$, $0 \le v \le 1$, $0 \le u + v \le 1$, but you can use barycentric coordinates to describe any point on the plane. Conversely, $$\left\lbrace\begin{aligned} u &= \frac{ x_1 (y - y_3) + x (y_3 - y_1) + x_3 (y_1 - y) }{x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) } \\ v &= \frac{ x_1 (y_2 - y) + x_2 (y - y_1) + x (y_1 - y_2) }{x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) } \\ \end{aligned}\right.\tag{4}\label{NA4}$$
Note that a direction vector $\underline{d} = (u, v)$ in barycentric coordinates corresponds to direction vector $\vec{d} = (x, y)$ in Cartesian coordinates via $$\vec{d} = u \hat{u} + v \hat{v} \quad \iff \quad \left\lbrace\begin{aligned} x &= ( - u - v) x_1 + u x_2 + v x_3 \\ y &= ( - u - v) y_1 + u y_2 + v y_3 \end{aligned} \right. \tag{5}\label{NA5}$$ and conversely, $$\left\lbrace\begin{aligned} u &= \frac{ x (y_3 - y_1) - y (x_3 - x_1) }{x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) } \\ v &= \frac{ x (y_1 - y_2) - y (x_1 - x_2) }{x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) } \\ \end{aligned}\right.\tag{6}\label{NA6}$$
Back to the topic at hand.
Let's investigate the case where $x_1 = 0$, $y_1 = 0$, $x_2 = B$, $y_2 = 0$.
If we apply $\eqref{NA4}$, we get $$\left\lbrace\begin{aligned} u &= \frac{x y_3 - x_3 y}{B y_3} \\ v &= \frac{y}{y_3} \end{aligned}\right .$$
Let's assume $x_3 \to X_3$ and $y_3 \to Y_3$, and see how $x \to X$, $y \to Y$ when barycentric coordinates $u$ and $v$ stay unchanged: $$\left\lbrace\begin{aligned} \frac{x y_3 - x_3 y}{B y_3} &= \frac{X Y_3 - X_3 Y}{B Y_3} \\ \frac{y}{y_3} &= \frac{Y}{Y_3} \\ \end{aligned}\right . \iff \left\lbrace\begin{aligned} x - x_3 \frac{y}{y_3} &= X - X_3 \frac{Y}{Y_3} \\ Y &= y \frac{Y_3}{y_3} \\ \end{aligned}\right .$$ which solving for $X$ and $Y$ yields $$\left\lbrace\begin{aligned} X &= x + (X_3 - x_3)\frac{y}{y_3} \\ Y &= y \frac{Y_3}{y_3} \\ \end{aligned}\right. \tag{7}\label{NA7}$$ So, this definitely fulfills OP's $y$ coordinate, but the $x$ coordinate is slightly different. Let's examine the shown diagram, and see how they differ.
In the diagram, $C = (x_3 , y_3) = (4.0, 8.0)$, $K = (x, y) = (4.0, 6.6)$, and $C1 = (X_3 , Y_3) = (7.9 , 7.0)$, and $K1 = (X , Y)$. Substituting into $\eqref{NA7}$ we get $K1 = (7.2, 5.8)$ which puts $K1$ slightly more left (on top of the vertical grid line immediately left of the red dot).
In the general case, we can calculate the direct mapping between point $(x,y)$ with respect to triangle $(x_1 , y_1)$, $(x_2 , y_2)$, $(x_3 , y_3)$, and point $(X,Y)$ with respect to triangle $(X_1 , Y_1)$, $(X_2 , Y_2)$, $(X_3 , Y_3)$, where both points have the same barycentric coordinates with respect to their own triangles, using $\eqref{NA4}$ and $\eqref{NA6}$.
By hand, that would be dull. But, if we construct the formulae in e.g. Maple, we get
$$\left\lbrace\begin{aligned} d &= x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2) \\ X_x &= ( X_1 (y_2 - y_3) + X_2 (y_3 - y_1) + X_3 (y_1 - y_2) ) \\ X_y &= ( x_1 (X_2 - X_3) + x_2 (X_3 - X_1) + x_3 (X_1 - X_2) ) \\ X_t &= X_1 (x_2 y_3 - x_3 y_2) + X_2 (x_3 y_1 - x_1 y_3) + X_3 (x_1 y_2 - x_2 y_1) \\ Y_x &= Y_1 (y_2 - y_3) + Y_2 (y_3 - y_1) + Y_3 (y_1 - y_2) \\ Y_y &= x_1 (Y_2 - Y_3) + x_2 (Y_3 - Y_1) + x_3 (Y_1 - Y_2) \\ Y_t &= Y_1 (x_2 y_3 - x_3 y_2) + Y_2 (x_3 y_1 - x_1 y_3) + Y_3 (x_1 y_2 - x_2 y_1) \end{aligned}\right .$$ so that $$\left\lbrace\begin{aligned} X &= \frac{x X_x + y X_y + X_t}{d} \\ Y &= \frac{x Y_x + y Y_y + Y_t}{d} \\ \end{aligned}\right.$$ Note that $d$ is twice the area of the first triangle, and can only be zero if the first triangle is a line or a point. The seven constants ($d$, $X_x$, $X_y$, $X_t$, $Y_x$, $Y_y$, and $Y_t$) only change when the coordinates of the first or second triangle change.
You can interpret the seven constants as a transformation, where $(X_t , Y_t)$ is the translation vector, $(X_x , X_y)$ is the new $x$ axis vector, and $(Y_x , Y_y)$ is the new $y$ axis vector, and $1/d$ is the scale factor.