**Polar Co-ordinates** are widely used in higher mathematics as well as in other branches of science. In polar co-ordinate system, the position of a point on the reference plane is uniquely determined referred to a fixed point on the plane and a half line drawn through the fixed point. The fixed point is called the Pole or Origin and the half line is drawn through the pole is called the Initial Line.

**Find the distance between two Points in Polar Co-ordinates**

**First Method: **Let OX be the initial line through the pole O of the polar system and (r₁, θ ₁) and (r₂, θ₂) the polar co-ordinates of the points P and Q respectively. Then, OP₁ = r₁, OQ = r₂, ∠XOP = θ₁ and ∠XOQ = θ₂, Therefore, ∠POQ = θ₂ – θ₁.

From triangle POQ we get,

PQ² = OP² + OQ² – 2 ∙ OP ∙ OQ ∙ cos∠POQ

= r₁² + r₂² – 2r₁ r₂ cos(θ₂ – θ₁)

Therefore, PQ = √[r₁² + r₂ ² – 2r₁ r₂ cos(θ₂ – θ₁)].

**Second Method:** Let us choose the origin and the positive x-axis of the cartesian system as the pole and initial line respectively of the polar system. If (x₁, y₁), (x₂, y₂) and (r₁, θ₁) (r₂, θ₂) be the respective Cartesian and polar co-ordinates of the points P and Q, then we shall have,

x₁ = y₁ cos θ₁, y₁ = r₁ sin θ₁

and, x₂ = r₂ cos θ₂, y₂ = r₂ sin θ₂.

Now, the distance between the points P and Q is

PQ = √[(x₂ – x₁)² + (y₂ – y₁)²]

= √[(r₂ cos θ₂ – r₁ cos θ₁)² + (r₂ sin θ₂ – r₂ sin θ₂)²]

= √[r₂² cos² θ₂ + r₁ ² cos² θ₁ – 2 r₁r₂ cos θ₁ cos θ₂ + r₂² sin² θ₂ + r₁²sin² θ₁ – 2 r₁r₁ sin θ₁ sin θ₂]

= √[r₂² + r₁² – 2r₁ r₂ Cos(θ₂ – θ₁)].

**Example on the distance between two points in polar Co-ordinates: **

Find the length of the line segment joining the points (4, 10°) and (2√3 ,40°).

**Solution:**

We know that the length of the line segment joining the points (r₁, θ₁), and (r₂, θ₂), is

√[ r₂² + r₁² – 2r₁ r₂ Cos(θ₂ – θ₁)].

Therefore, the length of the line-segment joining the given points

= √{(4² + (2√3)² – 2 ∙ 4 ∙ 2√(3) Cos(40 ° – 10°)}

= √(16 + 12 – 16√3 ∙ √3/2)

= √(28 – 24) = √4 = 2 units.

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