The lines $L$ and $K$ are symmetric to each other with respect to the line $y=x$. If the equation of the line $L$ is $y=ax+b$ with $a\neq 0$ and $b \neq 0$, then the equation of $K$ is $y=$

$\text{(A)}\ \frac 1ax+b \qquad \text{(B)}\ -\frac 1ax+b \qquad \text{(C)}\ \frac 1ax - \frac ba \qquad \text{(D)}\ \frac 1ax+\frac ba \qquad \text{(E)}\ \frac 1ax -\frac ba$
1. Given that the lines \( L \) and \( K \) are symmetric with respect to the line \( y = x \). This means that if the equation of line \( L \) is \( y = ax + b \), then the equation of line \( K \) can be found by interchanging \( x \) and \( y \) in the equation of \( L \).

2. Start with the equation of line \( L \):
   \[
   y = ax + b
   \]

3. To find the equation of line \( K \), interchange \( x \) and \( y \):
   \[
   x = ay + b
   \]

4. Solve for \( y \) in terms of \( x \):
   \[
   x = ay + b \implies ay = x - b \implies y = \frac{1}{a}x - \frac{b}{a}
   \]

5. Therefore, the equation of line \( K \) is:
   \[
   y = \frac{1}{a}x - \frac{b}{a}
   \]

The final answer is \(\boxed{\frac{1}{a}x - \frac{b}{a}}\)