If frac{1}{ab'} + frac{1}{ba'} = 0, then the | vedclass

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(C) The given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b'} + \frac{y}{a'} = 1$. Rewriting in slope-intercept form $y = mx + c

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Perpendicular to each other

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(C) The given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b'} + \frac{y}{a'} = 1$. Rewriting in slope-intercept form $y = mx + c$: $L_1: y = -\frac{b}{a}x + b$,so slope $m_1 = -\frac{b}{a}$. $L_2: y = -\frac{a'}{b'}x + a'$,so slope $m_2 = -\frac{a'}{b'}$. Two lines are perpendicular if $m_1 	imes m_2 = -1$. $m_1 	imes m_2 = (-\frac{b}{a}) 	imes (-\frac{a'}{b'}) = \frac{ba'}{ab'}$. Given condition: $\frac{1}{ab'} + \frac{1}{ba'} = 0 \implies \frac{ba' + ab'}{abb'a'} = 0 \implies ba' + ab' = 0 \implies ba' = -ab'$. Substituting this into the product of slopes: $m_1 	imes m_2 = \frac{-ab'}{ab'} = -1$. Since the product of slopes is $-1$,the lines are perpendicular.

If $\frac{1}{ab'} + \frac{1}{ba'} = 0,$ then the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b'} + \frac{y}{a'} = 1$ are

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