Question

The distance between the points (a cos   + b sin, 0) and (0, a sin  - b cos ), is:​

Answer 1

Let

• A = (a cosθ + b sinθ, 0)
• B = (0, a sinθ - b cosθ)

Applying distance formula,

$\boxed{\sf{AB = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} }}$

AB² = (x₂ - x₁)² + (y₂ - y₁)²

⇒ AB² = (0 - a cosθ - b sinθ)² + (a sinθ - b cosθ - 0)²

→ AB² = (-a cosθ - b sinθ)² + (a sinθ - b cosθ)²

➝ AB² = [(-a cos²θ) + (-b sinθ)² + 2(-a cosθ)(-b sinθ)]

          + (a²sin²θ + b²cos²θ - 2ab.sinθ.cosθ)

➝ AB² = a²cos²θ + b²sin²θ + 2ab.sinθ.cosθ + a²sin²θ + b²cos²θ - 2ab.sinθ.cosθ

Cancelling + 2ab.sinθ.cosθ and - 2ab.sinθ.cosθ,

AB² = a²cos²θ + b²sin²θ + a²sin²θ + b²cos²θ

→ AB² = a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

Since sin²θ + cos²θ = 1,

AB² = a² + b²

→ AB = √a² + b²

$\therefore \underline{\boxed{\bf{AB = \sqrt{a^{2} + b^{2} } }}}$