Point A exists on line between B and C points. I got "x,y" coordinates of B and C points, but only "y" of point A. How to find missing "x" of point A let's say that I have two points. Points B and C. I know their coordinates. Point B is 2,3 and point C is 4,6. I also know that there is point A located on the line between B and C. I know that it is "height", it's y coordinate, and it is 4. How can I learn the x of point A?
It's important for me to find the easiest way to calculate that - I feel like I am missing something obvious. I found other question like that, but it involved point A being outside of line between B and C. In my case, my point is directly on the line between these two points.
 A: Two points uniquely define a line, wich can be expressed by an equation $y=ax+b$. Try to find $a,b$ using the two points you alredy know, then solve the equation for $x$ using the value $y$ of A.
A: Here's a very useful way to parameterize the line $\overleftrightarrow{BC}$. Thinking of $B$ and $C$ as vectors, the points $A \in \overleftrightarrow{BC}$ are parameterized as
$$A = (1-t) B + t C, \quad t \in \mathbb R
$$
If you assign coordinates $A = (a_1,a_2)$, $B = (b_1,b_2)$ and $C = (c_1,c_2)$ then you can rewrite the parameterization equation in coordinates, and then use scalar multiplication and vector addition to rewrite this as two parameterization equations, one for $a_1$ and the other for $a_2$:
\begin{align*}
(a_1,a_2) &= (1-t)(b_1,b_2) + t (c_1,c_2) \\
(a_1,a_2)    &= ((1-t)b_1, (1-t)b_2) + (t c_1, t c_2) \\
(a_1,a_2)    &= ((1-t)b_1 + t c_1, (1-t)b_2 + t c_2) \\
a_1 &= (1-t)b_1 + t c_1 \\
a_2 &= (1-t)b_2 + t c_2 \\
\end{align*}
Now input what you know into the equations for $a_1$ and $a_2$, namely: $b_1=2$, $b_2=3$, $c_1=4$, $c_2=6$, $a_2=4$. Solve the $a_2$ equation for $t$, then plug that value of $t$ into the equation for $a_1$ to evaluate $a_1$.
Regarding your comments about whether $A$ is on the line outside of $B$ and $C$ or between $B$ and $C$, you can determine that by the value of $t$:

*

*If $t < 0$ then $A$ is outside beyond $B$.

*If $t=0$ then $A=B$.

*If $0 < t < 1$ then $A$ is between $B$ and $C$.

*If $t = 1$ then $A=C$.

*If $t>1$ then $A$ is outside beyond $C$.

However, the method for parameterizing $A$, and the method for finding the other coordinate of $A$ given one coordinate, is independent of where $A$ is positioned with respect to $B$ and $C$, requiring only that you know $A$ is somewhere on the line through $B$ and $C$.
A: $\vec{BA} = \lambda \vec{BC}$ where $0 \le \lambda \le 1$ since A is internal to the segment $BC$.
The above means
$$\lambda = \frac{x_A-x_B}{x_C-x_B}=\frac{y_A-y_B}{y_C-y_B}$$
You know the last term, then ..
A: 
It's important for me to find the easiest way to calculate that

With the given the information, the easiest way to determine $x_A$ is to equate the gradients of segments $BA$ and $BC$ (this is valid since $A,B,C$ are collinear): $$\frac{y_A-y_B}{x_A-x_B}=\frac{y_C-y_B}{x_C-x_B}\\\frac{y_A-y_B}{y_C-y_B}=\frac{x_A-x_B}{x_C-x_B}\\x_A=x_B+\frac{(x_C-x_B)(y_A-y_B)}{y_C-y_B}.$$
