Question

1. Find the distance between the following pair of points
(i) (2, 3) and (4, 1)
(ii)(-5,7)and(36,15)
(-5,7) and (-1,3)
(ii) (-2, -3) and (3,2)
(iv
)
(a, b) and (-a, -b)
re​

Answer 1

Solutions:-

(i) (2, 3) and (4, 1)

⟶ Here the given points are (2, 3) and (4, 1)

• Let A and B be the points on (2, 3) and (4, 1) respectively

Here,

The points become:-

• A(2, 3)
• B(4, 1)

Such that:-

x₁ = 2 and x₂ = 4

y₁ = 3 and y₂ = 1

We know,

Distance Formula = √(x₂ - x₁)² + (y₂ - y₁)²

Putting the values in distance formula:-

AB = √(4 - 2)² + (1 - 3)²

AB = √(2)² + (-2)²

AB = √4 + 4

AB = √8

AB = 2√3 units.

∴ The distance between A and B is 2√3 units.

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(ii) (-5, 7) and (36, 15)

⟶ Here the given points are (-5, 7) and (36, 15)

• Let P and Q be the points on (-5, 7) and (36, 15) respectively

Here,

The points become:-

• P(-5, 7)
• Q(36, 15)

Such that:-

x₁ = -5 and x₂ = 36

y₁ = 7 and y₂ = 15

Putting the values in distance formula:-

PQ = √[36 - (-5)]² + [15 - 7]²

PQ = √(36 + 5)² + (8)²

PQ = √(41)² + 64

PQ = √1681 + 64

PQ = √1745

⇒ PQ = 41.8 units.

∴ The distance between P and Q is √1745 (or 41.8) units.

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(iii) (-5, 7) and (-1, 3)

⟶ Here the given points are (-5, 7) and (-1, 3)

• Let R and S be the points on (-5, 7) and (-1, 3) respectively

Here,

The points become:-

• R(-5, 7)
• S(-1, 3)

Such that:-

x₁ = -5 and x₂ = -1

y₁ = 7 and y₂ = 3

Putting the values in distance formula:-

RS = √[-1 - (-5)]² + [3 - 7]²

RS = √(-1 + 5)² + (-4)²

RS = √(4)² + 16

RS = √16 + 16

RS = √32

RS = 4√2 units.

∴ The distance between R and S is 4√2 units

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(iv) (-2, -3) and (3, 2)

Here the given points are (-2, -3) and (3, 2)

• Let C and D be the points on (-2, -3) and (3, 2) respectively

Here,

The points become:-

• C(-2, -3)
• D(3, 2)

Such that:-

x₁ = -2 and x₂ = -3

y₁ = 3 and y₂ = 2

Putting the values in distance formula:-

CD = √[-3 - (-2)]² + [2 - 3]²

CD = √(-3 + 2)² + (-1)²

CD = √(-1)² + 1

CD = √1 + 1

CD = √2

∴ The distance between C and D is √2 units.

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(v) (a, b) and (-a, -b)

Here the given points are (a, b) and (-a, -b)

• Let E and F be the points on (a, b) and (-a, -b) respectively

Here,

The points become:-

• E(a, b)
• F(-a, -b)

Such that:-

x₁ = a and x₂ = -a

y₁ = b and y₂ = -b

Putting the values in distance formula:-

EF = √(-a - a)² + (-b - b)²

EF = √(-2a)² + (-2b)²

EF = √4a² + 4b²

EF = √4(a² + b²)

EF = 2√a² + b²

∴ The distance between E and F is 2√a² + b² units.

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