Question

The distance between the points (a cos \(\theta\) + b sin \(\theta\), 0) and (0, a sin \(\theta\) – b cos \(\theta\)) is

• A. \(\sqrt{a^2 + b^2}\)
• B. \(a^2 - b^2\)
• C. \(\sqrt{a^2 - b^2}\)
• D. \(a^2 + b^2\)

Hint:

For any points of the perpendicular form \((x, 0)\) and \((0, y)\), the distance is always \(\sqrt{x^2 + y^2}\).
Whenever you see terms like \((a\cos\theta + b\sin\theta)\) and \((a\sin\theta - b\cos\theta)\), their squares will always sum to \(a^2 + b^2\) because the cross-products cancel out and the remaining terms simplify using the identity \(\sin^2\theta + \cos^2\theta = 1\).
Recognizing this pattern helps you find the answer mentally!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic of this question is Coordinate Geometry combined with Trigonometric Identities.
We are given two points in the Cartesian coordinate plane.
The coordinates of these points are defined using trigonometric functions.
We need to find the straight-line distance between these two points and simplify the final algebraic expression.

Step 2: Key Formula or Approach:
- Use the standard distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] - We will also use the fundamental Pythagorean trigonometric identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \] - We will use algebraic expansion formulas:
\[ (x + y)^2 = x^2 + 2xy + y^2 \] \[ (x - y)^2 = x^2 - 2xy + y^2 \]

Step 3: Detailed Explanation:

• Identify the coordinates of the two given points:
\[ (x_1, y_1) = (a\cos\theta + b\sin\theta, 0) \] \[ (x_2, y_2) = (0, a\sin\theta - b\cos\theta) \]

• Substitute these coordinates into the distance squared equation \(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\):
\[ d^2 = (0 - (a\cos\theta + b\sin\theta))^2 + ((a\sin\theta - b\cos\theta) - 0)^2 \] \[ d^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 \]

• Expand the squared terms using algebraic identities:
\[ (a\cos\theta + b\sin\theta)^2 = a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta \] \[ (a\sin\theta - b\cos\theta)^2 = a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta \]

• Sum the two expanded terms to calculate \(d^2\):
Notice that the cross-product terms cancel each other out:
\[ +2ab\sin\theta\cos\theta - 2ab\sin\theta\cos\theta = 0 \] Therefore:
\[ d^2 = a^2\cos^2\theta + b^2\sin^2\theta + a^2\sin^2\theta + b^2\cos^2\theta \]

• Group the terms containing \(a^2\) and \(b^2\):
\[ d^2 = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta) \]

• Substitute the identity \(\sin^2\theta + \cos^2\theta = 1\):
\[ d^2 = a^2(1) + b^2(1) \] \[ d^2 = a^2 + b^2 \]

• Take the square root to determine the distance \(d\):
\[ d = \sqrt{a^2 + b^2} \]

Step 4: Final Answer:
The distance between the two points is \(\sqrt{a^2 + b^2}\).
Therefore, the correct option is (A).