Given two points $(x_1, y_1)$ and $(x_2, y_2)$, and a third point $(x_3,y_3)$ which is said to be equidistant from both the first two points, how would I find other points which are the same distance the third point is from the first two points.

Ex: $A = (7,3)$, $B = (6,0)$, and $C = (14,-1)$. $C$ is between $A$ and $B$. So the distance between the points is $\sqrt{65}$, so my question is asking how could I find another point which is this same distance between $A$ and $B$?

  • $\begingroup$ Hint: Think about the perpendicular bisector of a line. $\endgroup$ – stochasticboy321 Oct 29 '15 at 3:10
  • $\begingroup$ @stochasticboy321 What I had tried to do $√(7-x)^2+(3-y)^2$ = $√(6-x)^2+(0-4)^2$ I simplified this down to y = 11-x/3. I believe this will find any point that is equidistant between both the A and B I had specified, but this distance is not equal to $√65$. $\endgroup$ – user3657449 Oct 29 '15 at 3:14
  • $\begingroup$ You know that a point lies on a line, and you know it's distance from some given point. $(x - a_x)^2 + (y - a_y)^2 = d^2$ where $y = mx + c$. Substitute the latter into the former to get a quadratic in $x$. You'll find two solutions: The original point, and the point you're looking for. $\endgroup$ – stochasticboy321 Oct 29 '15 at 3:19
  • $\begingroup$ @stochasticboy321 Sorry i'm a bit confused by this explanation. I'm not sure how to apply it. $\endgroup$ – user3657449 Oct 29 '15 at 3:22

Geometrically speaking, all points that are equidistance from points, p and q, all lie on the perdendicular bisector of pq.

So if $p = (x1, y1)$ and $q =(x2, y2)$ are the points, the line pq has slope, $\frac {y2 - y1}{x2 - x1}$. So the slope of a perpendicular line will have slope m = $- \frac {x2 - x1}{y2 - y1}$ As the line is a bisector, it will contain the midpoint $mid = (\frac{x1+x2}{2},\frac{y1+y2}{2})$. So the perpendicular bisector is the line with the equation $y - \frac{y1+y2}{2} = m(x - \frac{x1+x2}{2}) = y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.

You want to find the points (x,y) where distance((x1,y1),(x,y)) = distance((x2,y2)(x,y)) = distance((x1,y1),(x2, y2)). Where (x,y) are on the line $y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.

So you have three equations:

$y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$

$\sqrt{(x1 - x)^2 + (y1 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$

and $\sqrt{(x2 - x)^2 + (y2 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$

There will be two possible points that satisfy the the equations.

Okay, that's hard. BUT if you are given one of the two points the other will be symmetrically placed on the other side of the midpoint.

Ex: A=(7,3), B=(6,0), and C=(14,−1).

So the midpoint of A and B is M = (6.5, 1.5). C is 14 - 6.5 = 7.5 away from M in the x direction. C is -1 - 1.5 = -2.5 away in the y directions. So the 4th point X will also be 7.5 away in the x direction and -2.5 away in the y direction. So X = (6.5 - 7.5, 1.5 - (-2.5)) = (-1, 4).


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