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I seem to have forgotten my coordinate geometry and this is absolutely blank.

Say I have $(x_1,y_1)$ and a distance $d$. Is it possible for me to solve for what the point $(x_2,y_2)$ is?

To explain more: If I have

$(x_1,y_1) = (1,2)$, and

$(x_2,y_2) = (3,4)$

Using the distance formula, I can calculate the distance to be $d=2\sqrt2$.

But if I have $(x_1,y_1) = (1,2)$ and $d=2\sqrt2$, how do I recover $(x_2,y_2)$ ?

Edit Yes, people are absolutely right when they say, multiple solutions are possible. Still, say, I also have an angle $\theta$ that tells me the orientation. Then, is it still not possible?

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  • $\begingroup$ You can't , just think about this, how many points are there of a distance 1 from the origin? $\endgroup$ – Quality Apr 2 '15 at 1:49
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No you don't have enough information to recover the $(x_2,y_2)$. The reason is that there are many points around $(x_1,y_1)$ which are a distance $d$ apart from it. In fact all points on the circle of radius $d$ centered at $(x_1,y_1)$ satisfy this. (This in fact the definition of a circle if you think about it.)

However, if you know that this point makes an angle $\theta$ from the $x$-axis, say, then yes you can recover $(x_2,y_2)$. The formula is $$x_2=x_1+d\times\cos\theta,\quad y_2=y_1+d\times\sin\theta,$$ where $d$ is the distance from $(x_1,x_2)$ to $(y_1,y_2)$.

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  • $\begingroup$ Right, but what if I also have an angle for that circle? $\endgroup$ – Raaj Apr 2 '15 at 1:51
  • $\begingroup$ @Raaj If you have the angle, then yes you can recover $(x_2,y_2)$. $\endgroup$ – Spenser Apr 2 '15 at 1:52
  • $\begingroup$ $dcos\theta$ and $dsin\theta$? $\endgroup$ – Raaj Apr 2 '15 at 1:54
  • $\begingroup$ @Raaj Can you clarify what do you mean by "$dcos\theta$ and $dsin\theta$?" I don't understand your question. $\endgroup$ – Spenser Apr 2 '15 at 1:57
  • $\begingroup$ No I just realized I could use those, and was making sure if that was it, and it looks like you answered around the same time :) $\endgroup$ – Raaj Apr 2 '15 at 2:05

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