How do you find the position of a point on a matrix I have a basic peice of algerbra I need to solve. Unfortunatly I didn't pay attention in high school.
I have three points two of which I know A and C.
C = 421,361
A = 50,50
B should be 50 units away on the diagonal line from A but still draw a straight line through C and A for example:

Any help is appreciated a formula or insight would be great, I have absolutely no idea how to go about this?
I apologize if my tags are inaccurate I know almost nothing about mathematics.
 A: The horizontal distance (along $x$) from $A$ to $C$ is $421-50$, which is $371$.
The vertical distance (along $y$) from $A$ to $C$ is $361-50$, which is $311$.
So, by Pythagoras' theorem, the diagonal distance is $\sqrt{371^2 +311^2}$. Doing the arithmetic, we get:
$$
\sqrt{371^2 +311^2} = \sqrt{137641 + 96721} = \sqrt{234362} = 484.1094917
$$
We only want to go 50 units along the diagonal, so we figure out what fraction of the diagonal distance this constitutes. We have
$$
\frac{50}{484.1094917} = 0.10328242
$$
So, we need to go $0.10328242$ of the way along the diagonal from $A$; call it $0.1033$, for short. To do this, we should go $0.1033$ of the $x$-distance, and $0.1033$ of the $y$-distance. 
So, our step in the $x$-direction should be $0.1033 \times 371$, which is $38.32$. Adding this to the $x$-coordinate of $A$, we get $50 + 38.32 = 88.32$. 
Similarly, our step in the $y$-direction should be $0.1033 \times 311$, which is $32.12$. Adding this to the $y$-coordinate of $A$, we get $50 + 32.12 = 82.12$. 
So, the desired point $B$ is at $x=88.32$, $y=82.12$ (roughly).
You could easily do all these calculations in Excel. Or, if you want code, it's as follows:
aX = 50   ;  aY = 50;
cX = 421  ;  cY = 361 ;

stepX = cX - aX;
stepY = cY - aY;

distance = sqrt(stepX*stepX + stepY*stepY);

fraction = 50/distance;

bX = aX + fraction*stepX;
bY = aY + fraction*stepY; 

A: Oh, OK, I didn't realize it was that specific. Take the vector from $A$ to $C$, so in your case, $(421-50,361-50) = (371,311)$. We want to go $50$ units along there, so we move $$50\cdot\frac{(371,311)}{\sqrt{371^2+311^2}}$$ from the point $A$. So the answer is that your point $B$ is located at $$(50,50) + 50\cdot\frac{(371,311)}{\sqrt{371^2+311^2}} = (50,50)+(38.318,32.121) = (88.318,82.121).$$
EDIT: To go one unit along the line segment from $A$ to $C$ we move $371/{\sqrt{371^2+311^2}} = .76636$ in the first coordinate and $311/{\sqrt{371^2+311^2}} = .64242$ in the second coordinate. Thus, stretching by a factor of $50$ gives $38.318$ in the first coordinate and $32.121$ in the second coordinate. Now displace each by $50$ (the coordinates of $A$).
