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# dr sarno back pain Equidistant Points – Definition, Theorem, Sphere, Construction …

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Equidistant points

From Latin: equi- “same”
Definition: A point P is equidistant from others if it is the same distance from them.
Try this Drag the point P. It will remain equidistant from Q and R.

A point is equidistant from other points if the
line segments
congruent (same length).
In the figure above as you move any of the three orange points and note
how P always remains the same distance (equidistant) from the other two.

## Alternate definition of a line

An alternate definition of a line is the “the set of all points equidistant from two given points”.
This line is known as the locus of the point P. In the figure above click “show locus” and
see that the green dotted line is the locus of all points that are equidistant from Q and R – a straight line.

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## Other point topics

• Point
• Plane
• Collinear points
• Equidistant points
• Intersection
• Vertex
• Midpoint
• Locus

# Equidistant Formula

Posted on by howtosolvemath

Equidistant means equal distance from every point. For example, consider the line segment containing the end points A and B and midpoint P. These  points are said to be equidistant if the distance between the point A and P is equal to the distance between the point P and B, that means the point P is the mid point of A and B. Therefore, we can call equidistant as mid point. It is shown in below figure.

## Equidistant formulas to find equidistant distance:

To find equidistant distance for any two end points, we have to use both mid point formula and distance formula. The midpoint formula and distance formula are given below.

Mid point formula = ((x1+x2) / 2, (y1+y2) / 2)

Distance formula = √((x2 – x1)2+(y2-y1)2)

Example:

Find the equidistant for the points A(8, 8) and B(4, 8) using above formulas.

Solution:

Here,  x1= 8

x2 = 4

y1= 8

y2 =  8

Step 1 : Find  mid point for the AB

AB = ((x1+x2 )/ 2,(y1+y2) / 2)

= ((8 + 4) / 2, (8 + 8) / 2)

= (12 / 2, 16 / 2)

Therefore, the mid  point of AB = (6, 8)

Consider P as mid point of AB.

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Step 2: Find distance between A(8, 8) and C(6, 8) using distance formula.

AC = √(x2 – x1)2 + (y2-y1)2

= √(6 – 8)2 + (8 – 8)2

= √(-2)2 + 0

= 2

Distance between AC  = 2

Step 3: Find distance between C(6, 8) and B(4, 8) using distance formula.

AC = √(x2 – x1)2 + (y2-y1)2

= √(4 – 6)2 + (8 – 8)2

= √(-2)2 + 0

= 2

Distance between CB  = 2

Here,

AC = CB

Therefore, AC and CB are equidistant.

## Example problem using equidistant formula :

Find x, if  the point A(x, 6) is equidistant from C(6, -4) and D(14, 8)

Solution:

Step 1: Find the distance between the points A(x, 6) and C(6, -4).

Distance, AC = √ ((6 – x)2 + (-4 -6)2)

= √ ((36 + x2 – 12x) + (-10)2 )

= √ (x2 – 12x + 136)

Step 2: Find the distance between the points A(x, 6) and D(14, 8).

Distance, A = √ ((14 – x)2 + (8 -6)2)

= √ ((196 + x2 – 28x) + 22 )

= √ (x2 – 28x + 200)

√ (x2 – 12x + 136) = √ (x2 – 28x + 200)

Squaring on both sides, we get

x2 – 12x + 136  = x2 – 28x + 200

– 12x + 28x = 200 – 136

16x = 64

x = 64 / 16

Therefore,      x = 4

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