Given, equation of lines are
$\begin{aligned} & \frac{x}{a}+\frac{y}{b} &=1 & \ldots \text { (i) } \\ \text { and } & \frac{x}{b}+\frac{y}{a} &=1 & \ldots \text { (ii) } \\ \Rightarrow & b x+a y &=a b & \ldots \text { (iii) } \\ \text { and } & a x+b y &=a b & \ldots \text { (iv) } \end{aligned}$
Solving Eqs. (iii) and (iv), we get
$$
\left(a^{2}-b^{2}\right) y=a^{2} b-a b^{2}=a b(a-b) \Rightarrow y=\frac{a b}{a+b}
$$
Substituting the value of $y$ in Eq. (iii), we get
$$
\begin{aligned}
& b x+a\left(\frac{a b}{a+b}\right)=a b \Rightarrow b x=a b-\frac{a^{2} b}{a+b} \\
\Rightarrow & b x=\frac{a b^{2}}{a+b} \Rightarrow x=\frac{a b}{a+b}
\end{aligned}
$$
$\therefore$ Point of intersection is $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$.
Since, equation of the line passing through origin is $y=m x$
$\therefore$ When it passes through $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$, then we get
$$
m=1
$$
Hence, requiret equation of line is $y-x=0$.