Question

find the the distance between the points (a cos theta,0) and (0,a sin theta)​

Answer 1

Step-by-step explanation:

Let P be the point (x₁, y₁) and Q the point (x₂, y₂) in a plane. Let the axes be at right angles to one another (rectangular). If d denotes the (shortest) distance (distance is always a positive quantity) between P and Q, we have

d = √{(x₁ - x₂)² + (y₁- y₂)²} ………………………………………………………………(1)

In the problem,

(x₁, y₁) = (cos θ, -sin θ), that is x₁ = cos θ and y₁ = -sin θ

(x₂, y₂) = (sin θ, cos θ), that is x₂ = sin θ and y₂ = cos θ

Substituting the above values of x₁, x₂, y₁ and y₂ in equation (1), we get

d = √{(cos θ - sin θ)² + (-sin θ - cos θ)²}

= √{(cos² θ-2cos θ.sin θ+sin² θ )+(sin θ+cos θ)²} [using formula (a-b)² = a² -2ab + b²]

= √(cos² θ-2sin θ cos θ+sin² θ+sin² θ+2sinθ cosθ+cos² θ)

[Using the formula (a+b)² = a² + 2ab + b²]

= √(cos² θ + sin² θ + sin² θ + cos² θ) [Cancelling the 2sinθ cosθ term ]

Now use the trigonometric identity cos² θ + sin² θ = 1.

∴ d = √(1+1) = √2

Hence the distance between the points (cos θ, -sin θ) and (sin θ, cos θ) is √2 .