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Q)

Find the equation of line parallel to $3x-4y+2=0$ and at a distance 2 from it?

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

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A)
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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