DISTANCE BETWEEN TWO POINTS

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the Cartesian plane (or xy –plane), at a distance ‘d’ apart. 

That is

d  =  PQ

By the definition of coordinates, 

OM  =  x1 ON  =  x2

MP  =  y1 NQ  =  y2

Step 1 :

Now, (PR ⊥ NQ)

PR  =  MN

(Opposite sides of the rectangle MNRP)

Step 2 :

Find the length of MN. 

MN  =  ON - OM

MN  =  x₂ - x₁

Step 3 :

Find the length of RQ. 

RQ  =  NQ - NR

RQ  =  y₂ - y₁

Step 4 : 

Triangle PQR is right angled at R. (PR ⊥ NQ)

By Pythagorean Theorem, 

PQ²  =  PR² + RQ² 

d²  =  (x₂ - x₁)² + (y₂ - y₁)²

By taking positive square root, 

d  =  √[(x₂ - x₁)² + (y₂ - y₁)²]

Given two points (x₁, y₁) and (x₂, y₂), the distance between these points is given by the formula

√[(x₂ - x₁)² + (y₂ - y₁)²]

Example 1 :

Find the distance between the two points given below. 

(-12, 3) and (2, 5)

Solution :

Formula for the distance between the two points is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Substitute (x₁, y₁)  =  (-12, 3) and (x₂, y₂)  =  (2, 5).

=  √[(2 + 12)² + (5 - 3)²]

=  √[14² + 2²]

=   √[196 + 4]

=  √200

=   √(2 ⋅ 10 ⋅ 10)

=   10√2

So, the distance between the given points is 10√2 units. 

Example 2 :

Find the distance between the two points given below. 

(-2, -3) and (6, -5)

Solution :

Formula for the distance between the two points is

√[(x₂ - x₁)² + (y₂ - y₁)²]

Substitute (x₁, y₁)  =  (-2, -3) and (x₂, y₂)  =  (6, -5).

=  √[(6 + 2)² + (-5 + 3)²]

=  √[8² + (-2)²]

=   √[64 + 4]

=  √68

=   √(2 ⋅ 2 ⋅ 17)

=   2√17

So, the distance between the given points is 2√17 units. 

Example 3 :

If the distance between the two points given below is 2√29, then find the value of k, given that k > 0.  

(-7, 2) and (3, k)

Solution :

Distance between the above two points  =  2√29

√[(x₂ - x₁)² + (y₂ - y₁)²]  =  2√29

Substitute (x₁, y₁)  =  (-7, 2) and (x₂, y₂)  =  (3, k).

√[(3 + 7)² + (k - 2)²]  =  2√29

√[10² + (k - 2)²]  =  2√29

√[100 + (k - 2)²]  =  2√29

Square both sides. 

100 + (k - 2)²  =  (2√29)²

100 + k² - 2(k)(2) +  2²  =  2²(√29)²

100 + k² - 4k +  4  =  4(29)

k² - 4k + 104  =  116

Subtract 116 from each side.

k² - 4k - 12  =  0

(k - 6)(k + 2)  =  0

k - 6  =  0  or  k + 2  =  0

k  =  6  or  k  =  -2

Because k > 0, we have

k  =  6

Example 4 :

Find the distance between the points A and B in the xy-pane shown below. 

Solution :

Identify the points A and B in the xy-plane above. 

Formula for the distance between the two points is

√[(x₂ - x₁)² + (y₂ - y₁)²]

To find the distance between the points A and B, substitute (x₁, y₁)  =  (2, -3) and (x₂, y₂)  =  (5, 5).  

AB  =  √[(5 - 2)² + (5 + 3)²]

AB  =  √[3² + 8²]

AB  =  √(9 + 64)

AB  =  √73 units

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.