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(A) Given lines are $ax + by + c = 0$ and $bx + ay + c = 0$.Subtracting the two equations: $(a - b)x + (b - a)y = 0 \Rightarrow (a - b)(x - y) = 0$.Si

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$a + b - c > 0$

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(A) Given lines are $ax + by + c = 0$ and $bx + ay + c = 0$.Subtracting the two equations: $(a - b)x + (b - a)y = 0 \Rightarrow (a - b)(x - y) = 0$.Since $a > b$,we have $a - b 
eq 0$,so $x = y$.Substituting $x = y$ into the first equation: $ax + bx + c = 0$ $\Rightarrow x(a + b) = -c$ $\Rightarrow x = \frac{-c}{a + b}$.Thus,the point of intersection is $P = \left(\frac{-c}{a + b}, \frac{-c}{a + b}ight)$.The distance between $(1, 1)$ and $P$ is given by $\sqrt{(1 - (\frac{-c}{a + b}))^2 + (1 - (\frac{-c}{a + b}))^2} $\sqrt{2(1 + \frac{c}{a + b})^2} $1 + \frac{c}{a + b} This implies $a + b - c > 0$.

For $a > b > c > 0$,the distance between $(1,1)$ and the point of intersection of the lines $ax + by + c = 0$ and $bx + ay + c = 0$ is less than $2\sqrt{2}$. Then:

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