If (x,y) and (X,Y) be the coordinates of the same point referred to two sets of rectangular axes with the same origin and if ax+by becomes pX+qY, where a,b are independent of x,y, then

Options

A. `a^(2)-b^(2)=p^(2)-q^(2)`
B. `a^(2)+b^(2)=p^(2)+q^(2)`
C. `a^(2)+p^(2)=b^(2)+q^(2)`
D. `a^(2)b^(2)=p^(2)q^(2)`

Explanation

To solve the problem, we need to establish the relationship between the coordinates (x, y) and (X, Y) in two different rectangular coordinate systems with the same origin. We are given that \( ax + by = pX + qY \), where \( a \) and \( b \) are constants independent of \( x \) and \( y \).

### Step-by-Step Solution:

1. **Establish the Relationship Between Coordinates:**
The coordinates of the same point in two systems can be expressed as:
\[ x = X \cos \theta - Y \sin \theta \]
\[ y = X \sin \theta + Y \cos \theta \]
Here, \( \theta \) is the angle of rotation between the two coordinate systems.

2. **Substitute the Relationships into the Given Equation:**
We substitute the expressions for \( x \) and \( y \) into the equation \( ax + by \):
\[ ax + by = a(X \cos \theta - Y \sin \theta) + b(X \sin \theta + Y \cos \theta) \]

3. **Distribute the Constants:**
Distributing \( a \) and \( b \) gives:
\[ ax + by = aX \cos \theta - aY \sin \theta + bX \sin \theta + bY \cos \theta \]

4. **Combine Like Terms:**
Now, we can group the terms involving \( X \) and \( Y \):
\[ ax + by = (a \cos \theta + b \sin \theta)X + (b \cos \theta - a \sin \theta)Y \]

5. **Identify the Coefficients:**
From the equation \( ax + by = pX + qY \), we can identify:
\[ p = a \cos \theta + b \sin \theta \]
\[ q = b \cos \theta - a \sin \theta \]

6. **Calculate \( p^2 + q^2 \):**
Now, we will compute \( p^2 + q^2 \):
\[ p^2 + q^2 = (a \cos \theta + b \sin \theta)^2 + (b \cos \theta - a \sin \theta)^2 \]

7. **Expand the Squares:**
Expanding both squares:
\[ p^2 = a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta \]
\[ q^2 = b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta \]

8. **Combine \( p^2 \) and \( q^2 \):**
Adding \( p^2 \) and \( q^2 \):
\[ p^2 + q^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta + b^2 \cos^2 \theta + a^2 \sin^2 \theta) \]
\[ = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\cos^2 \theta + \sin^2 \theta) \]

9. **Use the Pythagorean Identity:**
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ p^2 + q^2 = a^2 + b^2 \]

10. **Conclusion:**
Thus, we have shown that:
\[ p^2 + q^2 = a^2 + b^2 \]

### Final Answer:
The relationship is \( p^2 + q^2 = a^2 + b^2 \).