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(A) Given lines are $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$ and $\frac{x}{b} + \frac{y}{a} = 1$ $(2)$.Adding $(1)$ and $(2)$: $x(\frac{1}{a} + \frac{1}{

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$a^2y - b^2x = ab(a - b)$

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(A) Given lines are $\frac{x}{a} + \frac{y}{b} = 1$ $(1)$ and $\frac{x}{b} + \frac{y}{a} = 1$ $(2)$.Adding $(1)$ and $(2)$: $x(\frac{1}{a} + \frac{1}{b}) + y(\frac{1}{a} + \frac{1}{b}) = 2 \implies (x + y)(\frac{a + b}{ab}) = 2 \implies x + y = \frac{2ab}{a + b}$.Subtracting $(2)$ from $(1)$: $x(\frac{1}{a} - \frac{1}{b}) + y(\frac{1}{b} - \frac{1}{a}) = 0 \implies x(\frac{b - a}{ab}) - y(\frac{b - a}{ab}) = 0 \implies x = y$.Substituting $x = y$ in $x + y = \frac{2ab}{a + b}$,we get $2x = \frac{2ab}{a + b} \implies x = \frac{ab}{a + b}$.So,the intersection point is $P = (\frac{ab}{a + b}, \frac{ab}{a + b})$.The line joins $P$ and $Q = (a, b)$. The slope $m = \frac{b - \frac{ab}{a + b}}{a - \frac{ab}{a + b}} = \frac{\frac{ab + b^2 - ab}{a + b}}{\frac{a^2 + ab - ab}{a + b}} = \frac{b^2}{a^2}$.The equation is $y - b = \frac{b^2}{a^2}(x - a) \implies a^2y - a^2b = b^2x - ab^2 \implies a^2y - b^2x = a^2b - ab^2 = ab(a - b)$.

The equation of the straight line joining the point $(a, b)$ to the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b} + \frac{y}{a} = 1$ is

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