$\begin{array}{1 1}(a)\;3x-4y+12=0,3x-4y-8=0\\(b)\;3x-4y-8=0,3x-4y-8=0\\(c)\;3x-4y+10=0,3x-4y+8=0\\(d)\;3x+4y+12=0,3x+4y-8=0\end{array}$

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For a line to be parallel to line $3x-4y+2=0$ it should have the same slope

$\therefore $ The required line will be $3x-4y+c=0$

Now,

Distance between two parallel lines is $\large\frac{\mid c_1-c\mid}{\sqrt{a^2+b^2}}$

Hence $2=\large\frac{\mid 2-c\mid}{\sqrt{3^2+4^2}}$

$\therefore \pm =\large\frac{2-c}{5}$

Taking +ve sign

$10-2=-c$

$c=-8$

Taking -ve sign

$-10=2-c$

$c=12$

$\therefore$ Required equations are $3x-4y+12=0,3x-4y-8=0$

Hence (a) is the correct answer.

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