Explanation

(b)
Let the equation of $L_{1}$ be$y = mx$. Since, the intercepts made by the circle on $L_{1}$ and $L_{2}$ are equal, their distance from the centre of the circle are also equal
The centre of the given circle is $\left( \frac{1}{2},\ - \frac{3}{2} \right)$
$\therefore\ \left| \frac{\frac{1}{2} - \frac{3}{2} - 1}{\sqrt{1 + 1}} \right| = \left| \frac{m \times \frac{1}{2} + \frac{3}{2}}{\sqrt{m^{2} + 1}} \right|$
$\Rightarrow \ \frac{2}{\sqrt{2}} = \frac{|m + 3|}{2\sqrt{m^{2} + 1}}$
$\Rightarrow \ 8\left( m^{2} + 1 \right) = (m + 3)^{2}$
$\Rightarrow \ 7m^{2} - 6m - 1 = 0$
$\Rightarrow \ (m - 1)(7m + 1) = 0$
$\Rightarrow \ m = 1\ \ or\ \ m = - \frac{1}{7}$
So, the equations representing $L_{1}$ are
$y = x\ \ or\ \ \ y = \left( - \frac{1}{7} \right)x$
$\Rightarrow \ x - y = 0\ \$or$\ \ x + 7y = 0$