Find the distance between the parallel lines $3 x - 4 y + 7 = 0$ and $3 x - 4 y + 5 = 0$ .

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Solution

Given equations: $3 x - 4 y + 7 = 0$ and $3 x - 4 y + 5 = 0$
Comparing with $A x + B y + C_{1} = 0$ and $A x + B y + C_{2} = 0$ ,
We get $A = 3, B = - 4, C_{1} = 7, C_{2} = 5$
$\Rightarrow d = \frac{| C_{1} - C_{2} |}{\sqrt{a^{2} + b^{2}}}$
$\Rightarrow d = \frac{| 7 - 5 |}{\sqrt{3^{2} + 4^{2}}}$
$\Rightarrow d = \frac{2}{\sqrt{16 + 9}}$
$∴ d = \frac{2}{\sqrt{25}} = \frac{2}{5}$

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