On this image, we can know coordinate points PB and PA, by calculating vectors using formula: The line segment p1-p2 can be represented by the equation p = p1 + (p2 - p1)t, where t is a variable parameter. Varying t from 0 to 1 causes p to "sweep" from p1 to p2 along the straight line connecting the two points. The two special cases are t = 0 corresponding to p = p1, while t = 1 corresponds to p = p2.
This formula calculate length Pb and P2 equal P1-P2, but this isn't correct for me. I want to calculate Pb and Pa by using some variable parameter which set length Pb - P2 , and P2 - Pa
OR
I have another idea, to calculate triangle B point, if we know P1 - A (of triangle) P2 - C (of triangle) 90 degree corner of C and B, and B length. (Pb - Pa in the image)
Can any help with formula for calculate this?
$$\left\lbrace ~ \begin{aligned} \displaystyle x_a &= x_2 + \lambda_A (y_1 - y_2) \\ \displaystyle y_a &= y_2 + \lambda_A (x_2 - x_1) \\ \end{aligned} \right . , \quad \left\lbrace \begin{aligned} \displaystyle y_b &= x_2 - \lambda_B (y_1 - y_2) \\ \displaystyle y_b &= y_2 - \lambda_B (x_2 - x_1) \\ \end{aligned} \right .$$ where $\lambda_A$ is the length of $\overline{P2 ~ PA}$ compared to length $\overline{P1 ~P2}$, and $\lambda_B$ similarly for $\overline{P2 ~ PB}$.
Note the reversal of indexes for $y$ compared to $x$, this is intentional, because rotating $(x, y)$ 90 degrees counterclockwise yields $(-y, x)$. (Or $(y, -x)$ clockwise.)
If you want to use specific lengths, say $L_A$ and $L_B$, then use $$\left\lbrace ~ \begin{aligned} L_{12} &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \displaystyle \lambda_A &= \frac{L_A}{L_{12}} \\ \displaystyle \lambda_B &= \frac{L_B}{L_{12}} \\ \end{aligned} \right .$$