The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then `(1)/(a^(2))-(1)/(p^(2))+(1)/(b^(2))-(1)/(q^(2))` is equal to

Options

A. `-1`
B. 0
C. 1
D. none of these

Answer

To solve the problem, we need to analyze the line with intercepts \( a \) and \( b \) on the coordinate axes and how its intercepts change when the axes are rotated through a fixed angle.

### Step-by-Step Solution:

1. **Equation of the Line with Intercepts**:
The equation of the line \( L \) with intercepts \( a \) and \( b \) can be expressed as:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]

2. **Rotation of Axes**:
When the coordinate axes are rotated through an angle \( \theta \), the new coordinates \( (X, Y) \) are related to the old coordinates \( (x, y) \) by:
\[ x = X \cos \theta - Y \sin \theta \]
\[ y = X \sin \theta + Y \cos \theta \]

3. **Substituting into the Line Equation**:
Substitute the expressions for \( x \) and \( y \) into the line equation:
\[ \frac{X \cos \theta - Y \sin \theta}{a} + \frac{X \sin \theta + Y \cos \theta}{b} = 1 \]

4. **Rearranging the Equation**:
Multiply through by \( ab \) to eliminate the denominators:
\[ b(X \cos \theta - Y \sin \theta) + a(X \sin \theta + Y \cos \theta) = ab \]
This simplifies to:
\[ (b \cos \theta + a \sin \theta) X + (a \cos \theta - b \sin \theta) Y = ab \]

5. **Finding the New Intercepts \( p \) and \( q \)**:
The intercepts \( p \) and \( q \) on the new axes can be found by setting \( Y = 0 \) and \( X = 0 \) respectively:
- For \( Y = 0 \):
\[ (b \cos \theta + a \sin \theta) p = ab \implies p = \frac{ab}{b \cos \theta + a \sin \theta} \]
- For \( X = 0 \):
\[ (a \cos \theta - b \sin \theta) q = ab \implies q = \frac{ab}{a \cos \theta - b \sin \theta} \]

6. **Finding the Expression**:
We need to evaluate:
\[ \frac{1}{a^2} - \frac{1}{p^2} + \frac{1}{b^2} - \frac{1}{q^2} \]
Substitute \( p \) and \( q \):
\[ \frac{1}{p^2} = \frac{(b \cos \theta + a \sin \theta)^2}{a^2 b^2} \]
\[ \frac{1}{q^2} = \frac{(a \cos \theta - b \sin \theta)^2}{a^2 b^2} \]

7. **Combining the Terms**:
Substitute these into the expression:
\[ \frac{1}{a^2} - \frac{(b \cos \theta + a \sin \theta)^2}{a^2 b^2} + \frac{1}{b^2} - \frac{(a \cos \theta - b \sin \theta)^2}{a^2 b^2} \]
Combine the fractions:
\[ = \frac{b^2 - (b \cos \theta + a \sin \theta)^2 + a^2 - (a \cos \theta - b \sin \theta)^2}{a^2 b^2} \]

8. **Simplifying the Numerator**:
After simplifying the numerator, you will find that it equals zero:
\[ = 0 \]

### Final Result:
Thus, the expression evaluates to:
\[ \frac{1}{a^2} - \frac{1}{p^2} + \frac{1}{b^2} - \frac{1}{q^2} = 0 \]