The co-ordinate axes are rotated about the origin O in the counter-clockwise direction through an angle `60^(@)` If p and q are the intercepts made on the new axes by a straight line whose equation referred to the original axes is x + y = 1 , then `(1)/(p^(2)) + (1)/(q^(2)) = `

Options

A. 2
B. 3
C. 6
D. 8

Explanation

To solve the problem, we need to find the values of \( p \) and \( q \) after rotating the coordinate axes by \( 60^\circ \) and then calculate \( \frac{1}{p^2} + \frac{1}{q^2} \).

### Step-by-step Solution:

1. **Identify the Original Line Equation**:
The given line equation is:
\[ x + y = 1 \]

2. **Rotation of Axes**:
When the axes are rotated by an angle \( \theta \) (in this case \( 60^\circ \)), the transformation from the old coordinates \( (x, y) \) to the new coordinates \( (X, Y) \) is given by:
\[ X = x \cos \theta - y \sin \theta \]
\[ Y = x \sin \theta + y \cos \theta \]
For \( \theta = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \]

3. **Substituting Values**:
Substitute \( \cos 60^\circ \) and \( \sin 60^\circ \) into the transformation equations:
\[ X = x \cdot \frac{1}{2} - y \cdot \frac{\sqrt{3}}{2} \]
\[ Y = x \cdot \frac{\sqrt{3}}{2} + y \cdot \frac{1}{2} \]

4. **Express Old Coordinates in Terms of New Coordinates**:
Rearranging the equations gives:
\[ x = 2X + \sqrt{3}Y \]
\[ y = \sqrt{3}X - 2Y \]

5. **Substituting into the Original Line Equation**:
Substitute \( x \) and \( y \) into the line equation \( x + y = 1 \):
\[ (2X + \sqrt{3}Y) + (\sqrt{3}X - 2Y) = 1 \]
Simplifying this:
\[ (2 + \sqrt{3})X + (\sqrt{3} - 2)Y = 1 \]

6. **Finding Intercepts**:
To find the intercepts \( p \) and \( q \), we set \( Y = 0 \) to find \( p \):
\[ (2 + \sqrt{3})X = 1 \implies X = \frac{1}{2 + \sqrt{3}} \implies p = \frac{1}{2 + \sqrt{3}} \]
Now set \( X = 0 \) to find \( q \):
\[ (\sqrt{3} - 2)Y = 1 \implies Y = \frac{1}{\sqrt{3} - 2} \implies q = \frac{1}{\sqrt{3} - 2} \]

7. **Calculating \( \frac{1}{p^2} + \frac{1}{q^2} \)**:
First, calculate \( p^2 \) and \( q^2 \):
\[ p^2 = \left(\frac{1}{2 + \sqrt{3}}\right)^2 = \frac{1}{(2 + \sqrt{3})^2} = \frac{1}{4 + 4\sqrt{3} + 3} = \frac{1}{7 + 4\sqrt{3}} \]
\[ q^2 = \left(\frac{1}{\sqrt{3} - 2}\right)^2 = \frac{1}{(\sqrt{3} - 2)^2} = \frac{1}{3 - 4\sqrt{3} + 4} = \frac{1}{7 - 4\sqrt{3}} \]

Now, calculate \( \frac{1}{p^2} + \frac{1}{q^2} \):
\[ \frac{1}{p^2} = 7 + 4\sqrt{3}, \quad \frac{1}{q^2} = 7 - 4\sqrt{3} \]
Adding these gives:
\[ \frac{1}{p^2} + \frac{1}{q^2} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]

8. **Final Result**:
Therefore, the final answer is:
\[ \frac{1}{p^2} + \frac{1}{q^2} = 14 \]