In this mini-lesson, we will explore the world of equidistant. We will do so by learning the definition of equidistant, its general formula, how to calculate it, the role of the midpoint, and other interesting facts around the concept.

If we take points P and R, then from the below diagram, we can say that point Q is equidistant from two points P and R.

point Q is equidistant from point P and point R

Before we get started, did you know, the first use of the concept "Equidistant" was used in the year, 1556?

Actually, this word is derived from late Latin - aequi + distans meaning "of equal distance" from two points.

Let's move on and learn more about equidistant!

Lesson Plan

What Is the Definition of Equidistant?

A point is said to be equidistant from other points if it is always distanced equally or of the same distance away from each point.

Equidistant meaning in maths is mostly used in geometry, in the concepts of parallel lines/planes, perpendicular bisectors, circles, angle bisectors, etc.

In the illustration shown earlier, we can also say that Point Q is the midpoint to points P and R.

What Is the General Formula of Equidistant?

Generally, to find the equidistance for any two objects or points, we use both the midpoint formula and distance formula 

Midpoint Formula

When two points A (\(a\), \(b\)) and B (\(p\), \(q\)) are given, the midpoint, which is equidistant from each end will be

\[M = (\frac {a+p}{2}, \frac {b+q}{2})\]

Distance Formula

Considering the same points as above, the distance, d  between these two points will be

\[ d = \sqrt{(p - a^2) + (q - b)^2 } \]

Have a look at the equidistant calculator here: 

Try entering any two points and see what trajectory do its equidistant points show!


Thinking out of the box
Think Tank

Find the \(x\), if point A (x, 8) is equidistant from point C (6, 4) and B (12,12)

Is it possible to get 2 such points A? Can you justify your statement?

Solved Examples

Example 1



Find a point such that it lies on the y-axis and is equidistant from the points \((-1, 2\)) and \((2, 3)\)


We know that the x-coordinate of any point on the y-axis is \(0\)

Hence, we assume the point that is equidistant from the given points to be \((0,  k\)), i.e.

Distance between \((0, k)\) and \((-1, 2)\) \(=\) Distance between \((0, k)\) and \((2, 3)\)

\[\begin{aligned} \!\!\sqrt{(\!\!-\!1\!-\!0\!)^2+\!(\!2\!-\!k\!)^2}\!=\! \sqrt{(\!2-0\!)^2+(\!3-k\!)^2}\\ \end{aligned}\]

Squaring on both sides

\[ \begin{aligned}(-1-0)^2+(2-k)^2 &=(2-0)^2+(3-k)^2 \\1 + 4+k^2-4k &= 4+ 9 +k^2-6k \\2k &=8 \\ k&=4 \end{aligned}\]

Therefore, the required point is, \((0, k)=(0, 4)\)

\( \therefore\) Required Point  \(= (0, 4)\)
Example 2



The diameter of a circle has endpoints, (2, -3) and (-6, 5). 

Find the center of a circle.

Can you find the coordinates of the center of this circle?


The center of a circle divides the diameter into 2 equal parts.

So, it is a midpoint of the diameter.

Let \(x_{1}=2\), \(y_{1}=-3\), \(x_{2}=-6\), and \(y_{2}=5\)

The coordinates of the center is calculated as:

\[\begin{align}\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)&=\left(\dfrac{2+(-6)}{2}, \dfrac{-3+5}{2}\right)\\&=\left(-\dfrac{4}{2}, \dfrac{2}{2}\right)\\&=\left(-2,1\right)\end{align}\]

\(\therefore\) The center of the circle is \((-2,1)\)
Example 3



Consider the line segment \(\overline{AB}\) shown below. 

Find the midpoint of line segment

The points \(A=(1, h)\) and \(B=(5, 7)\) are equidistant from their midpoint M.

Find the value of \(h\) if the midpoint of \(\overline{AB}\) is \((3, -2)\)


Let \(x_{1}=1\), \(y_{1}=h\), \(x_{2}=5\), and \(y_{2}=7\)

According to the definition of midpoint we have,

\[\begin{align}\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)&=(3,-2)\\\left(\dfrac{1+5}{2}, \dfrac{h+7}{2}\right)&=(3,-2)\\\dfrac{h+7}{2}&=-2\\h+7&=-4\\h&=-11\end{align}\]

\(\therefore\) The value of \(h=-11\)
Example 4



The town of Lotto is mapped on a coordinate plane as shown.

The map of town of Lotto

The chocolate house is located at the point (3, 7) and the cake factory is located at the point (-1, -3)

The main entrance of the town is located on the x-axis but is at equal distance from the chocolate house and the cake factory.

Can you calculate the point of the main entrance?


To calculate the point of the main entrance, we have to calculate the midpoint of (3, 7) and (-1, -3)

Let \(x_{1}=3\), \(y_{1}=7\), \(x_{2}=-1\), and \(y_{2}=-3\).

The coordinates of the main entrance is calculated as:

\[\begin{align}\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)&=\left(\dfrac{3+(-1)}{2}, \dfrac{7+(-3)}{2}\right)\\&=\left(\dfrac{2}{2}, \dfrac{4}{2}\right)\\&=\left(1, 2\right)\end{align}\]

\(\therefore\) The main entrance is located at the point (1, 2)
important notes to remember
Important Notes
  1. The two or more objects or points are said to be equidistant if they all are equally distanced from each other. 
  2. A midpoint is also equidistant from the two original points. 
  3. We can use either the midpoint or distance formula to show that a point is equidistant from the two given points.

Interactive Questions

Here are a few activities for you to practice.

Select/Type your answer and click the "Check Answer" button to see the result.



Let's Summarize

We hope you enjoyed learning about equidistant with interactive questions. Now, you will be able to easily solve problems on equidistant formulas, graphs, and other related concepts of equidistant.  

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Frequently Asked Questions (FAQs)

1. What does it mean if two points are equidistant?

The two or more objects or points are said to be equidistant if they all are equally distanced from each other.

2. How do you find equidistant from 3 points?

In order to find a point that is equidistant from 3 different points, find the equation of the points equidistant from any 2 given points. 

Now, find the equation of the points equidistant from any one of the 2 previously chosen points, and the last remaining of the 3 given points. 

Solve the values you get on solving these two different equations. This final value is your answer.

3. How do you prove a point is equidistant?

To prove that a point is equidistant, simply equate the modulus of the distances between the given points and the equidistant point.

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