Calculate third point of triangle from two points and vector.

Question

I have $3$ points and their coordinates $A(a_x,a_y)$, $B(b_x,b_y)$ and $C(c_x,c_y)$. Then I have new coordinates of points $A(a_x',a_y')$ and $B(b_x',b_y')$. How to calculate the coordinates of point $C$ knowing that the distance from point $A$ to point $C$ and the angle between line $AB$ and line $AC$ are the same. It must work as a formula in any program or on a sheet of paper.

Thanks a lot from any help.

Answer 1

If I understand you correctly, $||A-C||=||A'-C'||$ and the angle between AB and AC and the one between A'B' and A'C' are equal. If not, please correct me :)

Lets set $r=||A-C||$ as the distance between A and C and $\alpha$ as the given angle between AB and AC. Also, let $\beta$ be the angle between the line A'B' and the horizontal axis.

Its easiest to solve this problem with vectors. The idea is to start from point B' in a direction that is given to you by the angle $\alpha$ and with the distance $r$.

If you know the angle between two lines, you can find the vector that goes into the direction (the correction with $\beta$ is not necessary if you align your coordinate system properly) as

$$v_1=\begin{pmatrix} \sin(\beta-\alpha) \\ \cos(\beta-\alpha)\end{pmatrix}$$

So to find your point C', compute $$C'= B'+r\begin{pmatrix} \sin(\beta-\alpha) \\ \cos(\beta-\alpha)\end{pmatrix}$$

Answer 2

Simple approach:

If you know the points then you can calculate the slope between $AB$ and $AC$ as $m_1$ and between $BA$ and $BC$ as $m_2$. Now as you have the slopes $m_1$ and $m_2$ you can take new line as reference $A'B'$ and make a line with $A'B'$ passing through $A'$ having slope $m_1$ similarly with $B'$ having slope $m_2$ , the intersection point of these lines should give you $C'$.