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(A) The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For the

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$20\sqrt{2}\beta = 11\alpha$

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(A) The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For the first pair of lines: $-x + y - 2 = 0$ and $x - y - 2 = 0$. Rewriting the first as $x - y + 2 = 0$,the distance $\alpha = \frac{|2 - (-2)|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.For the second pair of lines: $4x - 3y - 5 = 0$ and $8x - 6y + 1 = 0$. Dividing the second equation by $2$,we get $4x - 3y + 0.5 = 0$. The distance $\beta = \frac{|-5 - 0.5|}{\sqrt{4^2 + (-3)^2}} = \frac{5.5}{5} = 1.1 = \frac{11}{10}$.Now,calculating the ratio: $\frac{\alpha}{\beta} = \frac{2\sqrt{2}}{11/10} = \frac{20\sqrt{2}}{11}$.Thus,$11\alpha = 20\sqrt{2}\beta$.

Let $\alpha$ be the distance between the lines $-x + y = 2$ and $x - y = 2$,and $\beta$ be the distance between the lines $4x - 3y = 5$ and $6y - 8x = 1$,then:

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