distance between points

Question

If distance between the two points is 2 units ,then find a

(a,1),(2, 3)

Answer 1

Given two points are (a ,1) and (2 , 3)

Given that distance between two points is 2 units.

(x1 , y1) = (a , 1) and (x2 , y2) = (2 , 3)

Formula for distance between two points is

Substitute points

a = 2

Answer 2

Lets

A(x₁,y₁)=(a,1)

B(x₂,y₂)=(2,3)

Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by

d=√(x₂-x₁)²+(y₂-y₁)²

  =√(2-a)²+(3-1)²

  =√(2)²-2(2)(a)+a²+(2)²                           {(a-b)²=a²-2ab+b²}

  =√4-4a+a²+4

d=√a²-4a+8

The given problem distance between two points is 2 units

Then d=2

√a²-4a+8 = 2

Take squares each side we get

(√a²-4a+8)² = (2)²

Then

a²-4a+8 = 4

Subtract 4 each side we get

a²-4a+8-4 = 4-4

a²-4a+4 = 0                                                    (a²-4a+4=a²-2(a)(2)+(2)²=(a-2)²)

(a-2)² = 0

Take root each side we get

√(a-2)² = √0  

a-2 = 0

Adding each side by 2 we get

a-2+2=0+2

a=2

The value of a is 2.

Answer 3

Distance between two points = √(x2-x1)^2 +(y2-y1)^2

Given points say (x1,y1) =(a,1)

                         (x2,y2) = (2,3)

Substitute the points = √ [(2-a)^2+(3-1)^2

2 = √[(2^2+a^2-2*2*a)+(3^2+1^2-2*3*1)]

2 = √[4+a^2-4a+9+1-6]

2 = √(a^2-4a+8)

Sqare root each side

2^2 = (√(a^2-4a+8)^2

4 = a^2-4a+8

a^2-4a+8-4 = 0

0 = a^2-4a+4                              [(a-2)^2 = a^2-4a+4]

0 = (a-2)^2

a-2 = 0

⇒a = 2