FIND MISSING COORDINATE USING DISTANCE FORMULA

Find a given that :

Problem 1 :

P(2, 3) and Q(a, -1) are 4 units apart

Solution :

d = √(x₂ – x₁)² + (y₂ – y₁)²

Let x₁ = 2, y₁ = 3, x₂ = a, y₂ = -1

Here d = 4

4 = √(a – 2)² + (-1 – 3)²

4 = √(a – 2)² + (-4)²

4 = √(a – 2)² + 16

Taking square on both sides.

4² = (a – 2)² + 16

(a – 2)² = 0

a - 2 = 0

a = 2

So, the missing coordinates are a = 2. 

Problem 2 :

P(-1, 1) and Q(a, -2) are 5 units apart

Solution :

d = √((x₂ – x₁)² + (y₂ – y₁)²)

Let x₁ = -1, y₁ = 1, x₂ = a, y₂ = -2

Here d = 5

5 = √(a + 1)² + (-2 – 1)²

5 = √(a + 1)² + (-3)²

5 = √(a + 1)² + 9

Taking squares on both sides.

25 = (a + 1)² + 9

Subtracting 9 on both sides,

16 = (a + 1)²

Again take squares on both sides.

(a + 1) = √16

a + 1 = ± 4

So, the missing coordinates are a = 3 or a = -5. 

Problem 3 :

X (a, a) is √8 units from the origin

Solution :

X(a, a) and O(0, 0)

d = √((x₂ – x₁)² + (y₂ – y₁)²)

Let x₁ = a, y₁ = a, x₂ = 0, y₂ = 0

Here d = √8

√8 = √(0 - a)² + (0 – a)²

√8 = √a² + a²

√8 = √2a²

Taking squares on both sides.

8 = 2a²

8/2 = a²

4 = a²

a = ±2

Problem 4 :

A (0, a) is equidistant from P(3, -3) and  Q(-2, 2)

Solution :

AP = AQ

d = √((x₂ – x₁)² + (y₂ – y₁)²)

√((3 – 0)² + (-3 – a)²)= √((-2 – 0)² + (2 – a)²)

√(9 + (-3)² + a² – 2(-3)(a))) = √(4 + 2² + a² – 2(2)(a))

√(9 + 9 + a² + 6a) = √(4 + 4 + a² – 4a)

√18 + a² + 6a = √(8 + a² – 4a)

Taking square on both sides.

18 + a² + 6a = 8 + a² – 4a

18 + a² + 6a - 8 - a² + 4a = 0

10 + 10a = 0

10a = -10

a = -1

Problem 5 :

Find b given that A(3, -2) and B(b, 1) are √13 units apart.

Solution :

d = √((x₂ – x₁)² + (y₂ – y₁)²)

Let x₁ = 3, y₁ = -2, x₂ = b, y₂ = 1

Here d = √13

√13 = √((b – 3)² + (1 + 2)²)

√13  = √(b – 3)² + (3)²

√13  = √(b – 3)² + 9

Taking square on both sides.

13  = (b – 3)² + 9

Subtracting 9 on both sides.

13 - 9 = (b – 3)²

4 = (b – 3)²

b - 3 = √4

b - 3 = √4

b - 3 = ±2

So, the missing coordinates are b = 1 or b = 5.