Finding coordinate point if two points and time are known $A$ is at coordinate $(x_1,y_1)$ at $6:00$ am and is at coordinate $(x_2,y_2)$ at $6:15$ am. How can I know where $A$ will be at $6:05$ am i.e $(x_3,y_3)$? The value for $x_1=392,y_1=455,x_2=512,y_2=452$ are known and is straight line. 
I understand there is a formula to calculate this in coordinate mathematics. I will appreciate your answer and if you share the formula or term in mathematics.  
 A: It takes $A$ $15$ minutes, to go $-3$ units downward and $120$ units rightward. So he's traveling, $$\sqrt{3^2 + 120^2} = 3\sqrt{1601}$$ units per $15$ minutes.
Thus, at $6:05$, he'll have gone $\sqrt{1601}$, units from is starting position, $\left(\tfrac{1}{3}\text{ the distance }\right)$. We know that the point $(x_3,y_3)$ is on the line:
$$
y-455=\frac{-3}{120}(x-392)
$$
Thus, $y_3 = \frac{-3}{120}(x_3-392)+455$ 
AND:
$$
\sqrt{(x_3-x_1)^2+(y_3-y_1)^2} = \sqrt{(x_3-392)^2+(y_3-455)^2} = \sqrt{1601} 
$$
So, putting this all together we have:
$$
\sqrt{(x_3-392)^2+\left(\tfrac{-3}{120}(x_3-392)+455-455\right)^2} = \sqrt{1601} \Longrightarrow \dots
$$
$$
x_3=432, y_3 = 454
$$
EDIT
As Ross Millikan states, we can avoid the square roots entirely. See his comment below for the simplest solution to this problem. 
A: Is A moving at a constant speed? Use the distance formula: http://en.wikipedia.org/wiki/Distance_formula#Geometry
At a constant speed, the distance traveled is proportional to the time spent traveling because of the distance = rate x time equation.   
