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Equidistant Points

If you have ever looked at the street lights on your street carefully you would have noticed that they are equally spaced. That means the distance between two consecutive lights is pretty much the same along the street. Consider railway tracks. The distance between two consecutive sleepers of the tracks is same all along the tracks. That is exactly what 'equidistant' means. The distance between three or more points is same everywhere. The intersection points on the grid of your geometry paper, the corner points of any regular polygon, and the spoke end points of your bicycle wheel are all examples of equidistant points.

Graph

Equidistant

Bicycle Wheel

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Definition

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In a plane, if two or more points are such that they are at a fixed distance from each other or from another fixed point, then they are called equidistant points.
 
Equidistant points Defintion

In the above figure, the point E is equidistant from both A and B. Similar the point F is also equidistant from both A and B. Also the points G and H are both equidistant from the two points A and B. This is called being equidistant from two points.
 
Equidistant from both A and B

In the above figure, the points A, B, C and D are all equidistant from the center of the circle O. This is obvious since all the line segments OA, OB, OC, and OD represent the radii of the same circle and hence have to be equal. This is called being equidistant from a fixed point.

This concept of equidistant points is not limited to plane geometry only. In three-dimensional space also the concept of equidistant points exists. See the following tetrahedron as an example.
 
Equidistant points in three-dimensional space

All the five vertices of the tetrahedron, X, Y, Z, U, and V are equidistant from each other. This is the property of a regular tetrahedron. Let us look at another example.
 
Regular Tetrahedron

Every point on the sphere is equidistant from the center of the sphere that lies inside the sphere.

Theorem

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The set of equidistant points on a plane from a fixed point would form a circle with the fixed point as the center and the equidistance as the radius.
 Equidistant Theorem

In the above picture the fixed point is the center of the circle O. All the other points, A through L are equidistant from the fixed point O. Note that all the points between each of those points A and B, B and C and so on would also be equidistant from the center O, thus the set of all these points would form the circle with center O and radius equal to the equal distance OA or OB or OC or ..

How to find Equidistant points

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Given one point A, if we were to find two points B and C that lie at the same distance from A, then we can simply use the distance formula. Consider the coordinates of point A are ($a_1$, $a_2$) and those of B and C which we wish to find are ($b_1$, $b_2$) and ($c_1$, $c_2$). Then the distance AB can be given using the distance formula as:

$AB$ = $\sqrt{(b_1\ -\ a_1)^2\ +\ (b_2\ -\ a_2)^2)}$

Similarly, the distance AC can also be given using the distance formula as:

$AC$ = $\sqrt{(c_1\ -\ a_1)^2\ +\ (c_2\ -\ a_2)^2}$

Now, these two distances AB and AC have to be equal to each other. Thus,

$AB$ = $AC$

Let us say that they are equal to $r$. Then,

$AB$ = $AC$ = $r$

If the value of this r and the co-ordinates of point A are known to us then we can find the coordinates of $B$ and $C$. In fact, we can also find the equation of the circle containing all the points that are at a distance of $r$ from $A$. That equation would be:

$r^2$ = $(x\ -\ h)^2\ +\ (y\ -\ k)^2$

Here, (h,k) are the coordinates of the Center of the circle and (x,y) would represent any point on the circle. This ordered pair would definitely lie on the circle.

Equidistant points on a sphere

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A sphere can be thought of a set of many great circles as shown in the following figure:
 
Equidistant points on a sphere

There can be infinitely many great circles in a given sphere.
 
Great circles in a given sphere

Now, the center of each of those great circles would be same as the center of the sphere. In other words, all the great circles will have the same center.
 
Same center of the Great circles

All points on each great circle would be equidistant from the center of the circle. Since all the great circles have the same center, all points on all great circles would be equidistant from the center. In other words, all points on the sphere would be equidistant from the Center of the sphere.

How to Construct a Set of points that are Equidistant

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To construct a set of points that are equidistant from a given point on a plane is very easy. All we need to do is draw a circle with the given point as a center and the equidistance as the radius. We use a compass for the purpose. This is shown in the figure below:
 
Construct a set of points that are equidistant

To construct a set of points that are equidistant from two given points, we draw the perpendicular bisector of the line segment joining the two given points. All the points on this perpendicular bisector would be the points that are equidistant from the two given points. This can be done using two circular arcs of same radii drawn from both the given points as a Center. The line joining the point of intersection of these two arcs would be the required perpendicular bisector. This is shown below:

Perpendicular Bisector

Examples

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Example 1: Find an implicit equation of all points that are equidistant from the point (1, 2), the equal distance being $5$.

Solution:

The required set of points would represent a circle. The center of the circle is (1, 2) and 
the radius is $5$. Thus,

($h$, $k$) = (1, 2)

And

$r$ = 5

Substituting these into the general equation of the circle that we talked about earlier we 
have:

$5^2$ = $(x\ -\ 1)^2\ +\ (y\ -\ 2)^2$

$25$ = $(x\ -\ 1)^2\ +\ (y\ -\ 2)^2$ $\leftarrow$ Answer!

Example 2: Construct a set of points that are equidistant from the point ($-1$, $1$); the equal 
distance being $4$ cm.

Solution:

Take a compass and open it to a radius of $4$ cm. Then on your graph grid, plot the point 
($-1$, $1$). Then with this point as a center and the radius of 4cm on the compass, draw a 
circle.

Examples of Equidistant points
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