# Find which point is the closest to a specific point

Alright so I'm trying to find which points on a graph are closest to a specific point. The specific point is (4,4)... now we have 8 other random points on the graph, now we have to find which point is closer... here is a graph for a visual representation I understand you can just look at it and tell which point is closer but I'm looking for more of a formula.

Also, I would like to state that my highest mathematical level is pre-calculus so please bare that in mind when creating an answer (:>)

• If you know the collection of points and you know the specific point, all you would need to do is calculate each of those distances using the formula $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ and compare the distances. – Jonathan Davidson Jun 24 '17 at 4:23
• It's not clear if the question is asking for a systematic approach, in which case see the other comments and answers, or the particular case in the linked image. In the latter case, note that all points except $(5,2)$ are in different quadrants than $(4,4)$, so all the respective distances are $\gt 4$. But the distance between $(4,4)$ and $(5,2)$ is $\sqrt{(4-5)^2+(4-2)^2}=\sqrt{5} \lt 4\,$ – dxiv Jun 24 '17 at 4:32
• If you are as lazy as I am, if you just want to know what is the closest point $(x_i,y_i)$ to $(x_0,y_0)$ just compute $d^2_i=(x_i-x_0)^2+(y_i-y_0)^2$ and find the $i$ to which corresponds the minimum $d^2_i$. – Claude Leibovici Jun 24 '17 at 5:28

Euclidean Geometry tels us that the closest length between two point is a line. The fórmula between two point $x=(x_1,x_2)$ and $y=(y_1,y_2)$ on the Cartesian Plane $\mathbb{R}^2$ is $$d(x,y)=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}.$$ You can manually calculate the distance between $(4,4)$ and all the other eight point in the plane and compare which length is the smallest.