Euclidian Distance Formula

In mathematics, the Euclidian Distance Formula is used to measure the distance between two points. That distance is equal to the length of a straight line between the two given points.

It is calculated using the Pythagoras theorem with the help of the Cartesian coordinates of the points in a plane.

In the Euclidian plane, that is in two dimension, the distance between two points p(x1, y1) and q (x2,y2) will be:

d (x,y)=  √((x2- x1)2 + (y2– y1)2) 

In three dimension,

√((x2- x1)2 + (y2– y1)2+ (z2-z1)2) 

Solved Examples

Q1. Find the distance between points P (5, 2) and Q (4, 0).

Ans.                   d (x,y)=  √((x2- x1)2 + (y2– y1)2)

PQ = √((2-5)2 + (0-4)2)

PQ = √((-3)2 + (-4)2)

PQ =√(9 + 16)

PQ =√25

PQ =5 units

The Euclidian distance between points P (5, 2) and Q (4, 0) is 5 units.

Q2. Prove that the given points A (0, 4), B (6, 2), and C (9, 1) are collinear.

Ans. To prove that the given three points are collinear the sum of the distances between two pairs of the points should be equal to the distance between the third pair, this can be found through Euclid’s distance formula.

d (x,y)=  √((x2- x1)2 + (y2– y1)2) 

AB = √((6 – 0)2 + (2 – 4)2) = √(36 + 4) = √40 = 2√10

BC = √((9 – 6)2 + (1 – 2)2) = √(9 + 1) = √10 

CA = √((0 – 9)2 + (4 – 1)2) = √(81 + 9) = √90 = 3√10

Now since, 

2√10 + √10 = 3√10

It can be said that, AB + BC = CA

Hence proved that A, B, and C are collinear points.

Q3. The distance between two points A (5, – 3) and B (a, 1) is 5, find the value of a.

Ans. AB= √ (a-5)2+ (1-(-3))2

5= √ (a-5)2+ (1+3)2

5= √ (a-5)2+ (4)2

5= √ (a-5)2+16

Squaring on both sides,

(5)2 = (a-5)2-16

25-16= (a-5)2

9= a2-10a+25

a2-10a+25-9=0

a2-10a+16=0

a2-8a-2a+16=0                            {Replacing -10a by (-8a-2a)}

a(a-8)-2(a-8)=0

(a-2)(a-8)=0

Now, 

a-2=0, so a=2

a-8=0, so a=8

So, the values of a= 2, a=8.