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Q.
Let $a$ and $b$ be non-zero reals such that $a \neq b$. Then the equation of the line passing through the origin and the point of intersection of $\frac{x}{a} + \frac{y}{b} = 1 $ and $\frac{x}{b} + \frac{y}{a} = 1$ is
Given equation of lines are
$\frac{x}{a} + \frac{y}{b} = 1 $ ...(i)
and $ \frac{x}{b} + \frac{y}{a} = 1$...(ii)
$\Rightarrow \:\: bx +ay = ab$ ...(iii)
and $ax +by = ab$ ...(iv)
Solving (iii) and (iv), we get
$(a^2 - b^2)y = a^2 b -ab^2 =ab(a -b)$
$\Rightarrow \:\: y = \frac{ab}{a+b}$
Substituting the value of y in (iii}, we get
$bx+a\cdot \frac{ab}{a+b} =ab \Rightarrow bx =ab - \frac{a^{2}b}{a+b}$
$ \Rightarrow bx=\frac{ab^{2}}{a+b} \Rightarrow x =\frac{ab}{a+b}$
$\therefore $ Point of intersection is $\bigg( \frac{ab}{a+b} , \frac{ab}{a+b} \bigg)$
Since equation of the line passing through origin is $y = mx$
$\therefore $ When it pass through $\bigg( \frac{ab}{a+b} , \frac{ab}{a+b} \bigg)$ then, we get $m = 1$
Hence, required equation of line is,
$y - x = 0$