Question
Find the distance between the parallel lines 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0.
∴ Distance between the parallel lines
$\begin{aligned} & =\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right| \\ & =\left|\frac{3-15}{\sqrt{3^2+4^2}}\right|=\left|\frac{-12}{\sqrt{9+16}}\right|\end{aligned}$
$=\frac{12}{5}$ units
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$2 x^2-\sqrt{3} x+1=0$