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$A$ 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.

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In my answer, I assume $A$ moves at a constant speed. Is that what you need? – Rustyn Sep 24 '13 at 3:45

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.

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You needn't do the square roots. Just compute the $x$ velocity $v_x=\frac {-3}{15}$ and $y$ velocity $v_y=\frac {120}{15}=8$ The position at time $t$ (measured in minutes after $6:00$) is $(x_1+v_xt, y_1+v_yt)$ – Ross Millikan Sep 24 '13 at 3:48
@RossMillikan You are absolutely right. Except for I think you got them switched around? $v_x = 120/15$, $v_y = -3/15$ – Rustyn Sep 24 '13 at 3:49

Is A moving at a constant speed? Use the distance formula:

At a constant speed, the distance traveled is proportional to the time spent traveling because of the distance = rate x time equation.

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