Find y-coordinate on a line between two (known) points Iam a littlebit stuck with a simple task and hope to find some help here, since my days in school are now long time over and to be honest i can’t remember so well how to do it. 
I have a straight line between two points, lets say (8,20) (300,50) and i want to figure out whats the y-value of (200,y). Now i think i need to find the slope by using (y2-y1)/(x2-x1). But from there i’am stuck. 
Any help is appreciate
 A: Slope is a good way of doing it.  Somewhat fancier, but very useful, is the fact that the points on the line joining $(a,b)$ to $(c,d)$ are given by
$$(x,y)=\left((1-t)a+tc, (1-t)b+t d\right).\tag{1}$$
This is the parametric equation of the line. The obvious modification works equally well in three dimensions.
In your case, we have $(a,b)=(8,20)$ and $(c,d)=(300,50)$.  We know that $x=200$ and want to find $y$. If we can find $t$, we will be nearly finished.
Note that $(1-t)(8)+(t)(300)=200$. Solve for $t$.
A: The other solutions are perfectly correct, but seem unnecessarily algebraic to me.
I would solve the problem using proportional reasoning:
The horizontal distance between the two points you know is $292$ (which you can get by subtracting $300-8$).  You are looking for a point $192$ units from the left-hand point.  That's $192/292 \approx 65.75\%$ of the way from the right edge.
The vertical distance between the two points you know is $30$ (from $50-20$).  So the $y$-coordinate you are looking for is $65.75\%$ of $30$, or about $19.725$, units up from $20$.  So the answer is about $39.725$.
A: Compute the gradient,
$$ m= \frac{y_2-y_1}{x_2-x_1} $$
Then the y-intercept is
$$ c=y_1 - mx_1 $$
or you could do $c=y_2-mx_2$ which would give the same answer.
Then to find $y$ for a particular $x$ (e.g. $x=200$), you just do
$$y = mx+c$$
A: If you want to solve this using homogeneous coordinates then you establish the $(A,B,C)$ coefficients of the line forming the equation $Ax+By+C=0$ and then solve for $y$ when $x=200$.
The line coefficients are:
$$ \begin{pmatrix} A \\ B \\ C \end{pmatrix} = \begin{pmatrix} 8 \\ 20 \\ 1 \end{pmatrix} \times \begin{pmatrix} 300 \\ 50 \\ 1 \end{pmatrix} = \begin{pmatrix} -30 \\ 292 \\ -56000 \end{pmatrix}$$ where $\times$ is the vector cross product.
To find your point solve $$ \begin{pmatrix} 200 \\ y \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -30 \\ 292 \\ -56000 \end{pmatrix} =0$$ where $\cdot$ is the vector dot product.
$(200)(-3)+(y)(292)+(1)(-56000)=0 \} y = \frac{2900}{73} $
