Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x + y = 1.$ If the intercepts made by the circle $x^2 + y^2 - x + 3y = 0$ on $L_1$ and $L_2$ are equal, then which of the following equations can represent $L_1$?

The centre and radius of the given circle are $\left(\frac{1}{2},-\frac{3}{2}\right)$ and $\sqrt{\frac{5}{2}}$ respectively.

Let $y=mx$ be the equation of $L_1$ Then,

       $p_1=$ Length of the intercept on $L_1$

$\Rightarrow p_1=2\sqrt{{\left(\sqrt{\frac{5}{2}}\right)}^2-{\left(\frac{m+3}{2\sqrt{m^2+1}}\right)}^2}=2\sqrt{\frac{5}{2}-\frac{(m+3{)}^2}{4\left(m^2+1\right)}}$

and,

$\Rightarrow p_2=$ Length of the intercept on $L_2$

$\Rightarrow p_2=2\sqrt{{\left(\sqrt{\frac{5}{2}}\right)}^2-(\sqrt{2}{)}^2}=2\sqrt{\frac{5}{2}-2}=\sqrt{2}$

Now,

$\begin{array}{l} p_1=p_2\\ \Rightarrow 2\sqrt{\frac{5}{2}-\frac{(m+3{)}^2}{4\left(m^2+1\right)}}=\sqrt{2}\\ \Rightarrow 5-\frac{(m+3{)}^2}{4\left(m^2+1\right)}=1\\ \Rightarrow 7m^2-6m-1=0\end{array}$

$\Rightarrow (m-1)(7m+1)=0\Rightarrow m=1,-\frac{1}{7}$

So, the equations of $L_1$ are $y = x$ and $7y = - x.$