FIND MISSING COORDINATE USING DISTANCE FORMULA

Example 1 :

Find the value of a if the distance between the points at (7, 5) and (a, -3) is 10 units.

Solution :

d = √(x₂ - x₁)² + (x₂ - x₁)²

Let x₂ = a, x₁ = 7, y₂ = -3, y₁ = 5

Here d  =  10

10 = √(a - 7)² + (-3-5)²

10 = √(a - 7)² + (-8)²

10 = √(a - 7)² + 64

10 = √(a² - 2 a (7) + 7²) + 64

10 = √(a² - 14 a + 113)

Taking squares on both sides

10² = (√(a² - 14 a + 113)²

a² - 14 a + 113 - 100  =  0 

a² - 14 a + 13  =  0 

(a - 1) (a - 13)  =  0 

a  =  1  (or)  a  =  13

So, the missing coordinates are 1 or 13.

Example 2 :

Find the value of a if the distance between the points at (3, -1) and (a, 7) is 10 units.

Solution :

d = √(x₂ - x₁)² + (x₂ - x₁)²

Let x₂ = a, x₁ = 3, y₂ = 7, y₁ = -1

Here d  =  10

10 = √(a - 3)² + (7 - (-1))²

10 = √(a - 3)² + (7 + 1)²

10 = √(a - 3)² + (8)²

10 = √(a² - 2 a (3) + 3²) + 64

10 = √(a² - 6a + 9 + 64)

10 = √(a² - 6a + 73)

Taking squares on both sides

10² = (√(a² - 6 a + 73)²

a² - 6 a + 73 - 100  =  0 

a² - 6 a - 27  =  0 

(a - 9) (a + 3)  =  0 

a  =  9  (or)  a  =  -3

So, the missing coordinates are 9 or -3.

Example 3 :

Find the value of a if the distance between the points at (10, a) and (1, -6) is √145 units.

Solution :

d = √(x₂ - x₁)² + (x₂ - x₁)²

Let x₂ = 1, x₁ = 10, y₂ = -6, y₁ = a

Here d  =  √145

√145 = √(1 - 10)² + (-6 - a)²

√145 = √(- 9)² + (-6 - a)²

√145 = √81 + (6 + a)²

√145 = √81 + (6² + 2(6) a + a²)

√145 = √(81 + 36 + 12 a + a²)

√145 = √a² + 12a + 117

Taking squares on both sides

(√145)² = (√a² + 12a + 117)²

a² + 12 a + 117 - 145  =  0 

a² + 12 a - 28  =  0 

(a + 14) (a - 2)  =  0 

a  =  -14  (or)  a  =  2

So, the missing coordinates are -14 or 2.

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